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What is the limiting reactant in Mg plus HCl?
In the reaction between magnesium (Mg) and hydrochloric acid (HCl), the limiting reactant is the reactant that is consumed first, which determines the maximum amount of product that can be formed. To determine the limiting reactant, you would need to compare the moles of magnesium and hydrochloric acid to see which one is present in the lowest stoichiometric amount.
What is the limiting reactant in c6h6 plus hno3---c6h5no2 plus H2O?
The limiting reactant is the reactant that is completely consumed in a chemical reaction. In this case, you would need to compare the moles of each reactant to see which one is completely used up first. Whichever reactant is present in the lowest stoichiometric amount is the limiting reactant.
How do you know which reactant in a process is the limiting reactant?
The limiting reactant or reagent can be determined by calculating the number of moles of each reactant/reagent. Whichever is the lowest number of moles is the limiting reagent in the reaction, assuming that stoichiometry is 1;1
Conversions of stoichiometry?
Stoichiometry Conversions Using Balanced EquationsMol A --> Mol B (Mol -->Mol)("How many moles of B are needed to react with X mol A?") __A + __B --> __AB1. Multiply the number of moles (A) by the number of moles (# molecules)1 for B.2. Divide the result by the number of molecules for A.- FormulaMol A * # molecules B/# molecules A- Conversion factorMol A * # molecules (mol) B = Mol B----- # molecules (mol) A*****1 the number of moles in this case refers to however many molecules of each substance is within the balanced equation (# molecules). For clarity sake I put # molecules in place of "moles" where appropriate; however, on practice problems or demonstrations shown in textbooks, it's likely it will say moles instead of # molecules.*****Mol A --> Mass B (Mol --> Mass)("How many grams B are needed to produce X mol A?") __A + __B --> __AB1. Multiply mol A by # molecules B; divide by # molecules A. (Mol A --> Mol B)2. Multiply result from 1 by molar mass B (mol B --> Mass B).- FormulaMol A x # molecules B/# molecules A- Conversion FactorMol A x # molecules (mol) B x molar mass B = mass B--------- # molecules (mol) A ----- 1 mol BMass A --> Mol B (Mass --> Mol)("How many mol of B are needed to react with X g A?") __A + __B --> __AB1. Multiply Mass A by # molecules B.2. Divide by molar mass multiplied by # molecules A.- FormulaMass A x # molecules B/(Molar Mass A x # molecules A)- Conversion FactorMass A x 1 mol A x # molecules (mol) B = mol B--- molar mass A - # molecules (mol) AMass A --> Mass B (Mass --> Mol --> Mass)("How many grams of A can be produced from X grams B?") __A + __B --> __AB1. Convert from grams (g) to mol for substance A (mass A --> mol A).2. Divide mol A by mol B (mol A --> mol B [Molar Ratio of Substances]).3. Multiply mol B by molar mass B (mol --> mass)In summary, you are converting from grams A to mol A, then mol A to mol B, then mol B to grams B.- FormulaMass A x # molecules B x Molar Mass B/(Molar Mass A x # molecules A)- Conversion FactorMass (g) A * 1 mol A -- x -- # molecules (mol) B -- x -- molar mass B = mass B----- molar mass (g) A --x-- # molecules (mol) A --- x --- 1 mol BLimiting Reagent and Reagent in ExcessThe limiting reagent or limiting reactant is the substance that limits the reaction. ("What substance has the least amount produced from a reaction?"). The reagent in excess or reactant in excess is the product left over or reagent that is leftover from the reaction; in other words, the reagent that has the most product that did not react.To identify the limiting reagent or limiting reactant, identify which substance produced the least amount of product (which reactant yields the least amount of product).Steps (given masses of products):A. Identify amount of product created per reactant (reactant --> product yield).1. Balance the equation if it has not been done already.2. Convert the given masses of reactant (A, B, etc.) to mass product (C) (see "Mass A --> Mass B [Mass --> Mass]" above). (mass reagent --> mass product)B. Identify the limiting reagent. The limiting reagent (reactant) will be the reactant (A, B, etc.) that yields the least amount of C (product).C. Identify the reagent in excess. The reagent in excess (reactant) will be the reactant (A, B, etc.) that yields the most amount of C (product).D. Give how much reagent in excess remain unreacted. (How much reactant is leftover). For simplicity sakes, the limiting reagent will be A and the reagent in excess will be B. "For every X grams of the limiting reagent, there is Y grams of the reagent in excess".Mass Limiting Reactant --> Mass Reagent in Excess:First convert the mass of the limiting reagent to the mass of the reagent in excess (mass limiting reagent : mass reagent in excess [reacting] ratio); then subtract the mass of the limiting reagent from the mass of the reagent in excess (that reacted)1. Convert the mass of the limiting reagent to mass of the reagent in excess (ratio mass limiting reagent: mass reagent in excess). Refer to "Mass A --> Mass B" above.2. Subtract the original amount of B (reagent in excess) from the amount of B needed to react with A (limiting reagent).Reagent in Excess leftover = Starting Mass A - Reacting Mass B (step 1 answer)E. Find % yield.% yield = actual yield (given)------- theoretical yield (must be found)** the theoretical yield is the amount of product theoretically produced by the limiting reagent; the actual yield is the amount of product actually produced by the reactants; the theoretical yield will have been found in step A. The actual yield will be given within the worded problem.Ex. 4Na2CO3 + Fe3Br8 --> 8NaBr + 4CO2 + Fe3O4 A. How many grams of Fe3O4 can be produced from 100.0g Na2CO3 and 300.0g Fe3Br8?100.0g Na2CO3 x 1 mol Na2CO3 x 1 mol Fe3O4 x 231.6g Fe3O4------------------ 106.0g Na2CO3 -- 4 mol Na2CO3 -- 1 mol Fe3O4= 54.62g Fe3O42(2 this is known as the theoretical yield, which will be needed when calculating percentage yield later on).300.0g Fe3Br8 x 1 mol Na2CO3 x 1 molecules Fe3O4 x 231.6g Fe3O4----------------- 106.0g Fe3Br8 ---- 4 molecules Fe3Br8 -- 1 mol Fe3O4= 86.12g Fe3O4B. What is the limiting reagent?The limiting reagent in this case is Na2CO3 because it has the lowest theoretical yield for producing Fe3O4 (54.62g Fe3O4 vs 86.12g Fe3O4).C. What is the reagent in excess?The reagent in excess is Fe3Br8 because it has the highest theoretical yield for producing Fe3O4.D. How many grams of the reagent in excess(Fe3Br8)remain unreacted?100.0g Na2CO3 x 1 mol Na2CO3 x 1 molecules Fe3Br8 x 806.8g Fe3Br8------------------- 106.0g Na2CO3 -- 4 molecules Na2CO3 - 1 mol Fe3Br8= 190.3g Fe3Br8300.0g Fe3Br8 - 190.3g Fe3Br8 = 109.7g Fe3Br8 leftover (unreacted)E. If 42.75g of Fe3O4 were isolated, what is the % yield?% yield = 42.75g----------- 54.62g2 x 100% = 78.27%
How many grams carbon dioxide is made when 42 grams of propane is burned with 115 grams of oxygen how many grams of water is made?
First you need to write a balanced equation. You are given that propane undergoes a combustion reaction that produces carbon dioxide and water.Unbalanced: C3H8 + O2 ---> CO2 + H2OBalanced: C3H8 + 5O2 ---> 3CO2 + 4H2OGivens:42.0 grams C3H8 (Molecular mass 44.0 g)115.0 grams O2 (Molecular mass 32.0 g)Molecular mass of CO2: 44.0 gMole ratio 1:5:3:4 (C3H8:O2:CO2:H2O)Then you need to find which of the reactants are the limiting reactant (lowest value) and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.42.0 g C3H8 / (44.0 g) = .955 moles C3H8 115.0 g
What is the limiting reactant in Mg plus HCl?
In the reaction between magnesium (Mg) and hydrochloric acid (HCl), the limiting reactant is the reactant that is consumed first, which determines the maximum amount of product that can be formed. To determine the limiting reactant, you would need to compare the moles of magnesium and hydrochloric acid to see which one is present in the lowest stoichiometric amount.
What is the limiting reactant in c6h6 plus hno3---c6h5no2 plus H2O?
The limiting reactant is the reactant that is completely consumed in a chemical reaction. In this case, you would need to compare the moles of each reactant to see which one is completely used up first. Whichever reactant is present in the lowest stoichiometric amount is the limiting reactant.
What is the limiting reactant reagent and what is excess reagent in recrystallization?
The Limiting Reactant is the smaller number once you compare the two reactants with one product. The product that you are comparing them both with must be the same. The Excess Reactant is the larger number, or the amount left over in the chemical reaction.
How do you know which reactant in a process is the limiting reactant?
The limiting reactant or reagent can be determined by calculating the number of moles of each reactant/reagent. Whichever is the lowest number of moles is the limiting reagent in the reaction, assuming that stoichiometry is 1;1
How do you calculate theoretical yield?
When calculating the theoretical yield of a product within a reaction, the idea is to convert mass reactant --> mass product; in other words, "how many grams of product X will N grams reactant A and N grams reactant B yield (create)?" Whichever mass is the lowest is considered the theoretical yield (in other words, the reactant that produces the least amount of product).__A + __B --> __Xmass reactant A => mass product X (mass A: mass X)mass reactant B => mass product X (mass B: mass X)In order to calculate the theoretical yield, you need to convert the mass of reactant A to the mass of product X AND the mass of reactant B to the mass of product X.1. Multiply the mass of the reactant by the number of molecules (or moles) X and by the molar mass of X.2. Divide the above by the molar mass of reactant (which is multiplied by the number of molecules [or moles] of reactant).Do this for each reactant involved.- FormulaMass reactant * # molecules (moles) product X1 * molar mass product X--------- # molecules (moles) reactant * molar mass reactant= Mass Product X- Conversion FactorMass reactant * 1 mol reactant * # molecules product X * molar mass product X-------------- molar mass reactant - # molecules reactant --- 1 mol product X= Mass Product X1the # molecules is obtained from the balanced equation. For clarification, I stated # molecules in place of mol; however, on practice problems and in textbook tutorials on how to solve problems in regards to stoichiometry, it is likely it will say mol or moles instead of # molecules.Ex. Ca(NO3)2 + 2NaF --> CaF2 + 2NaNO3What is the theoretical yield of CaF2 (product X) when 43.5g Ca(NO3)2 (reactant A) react with 39.5g NaF (reactant B)?43.5g Ca(NO3)2 * 1 mol Ca(NO3)2 * 1 molecule1 CaF2 * 78.08g CaF2-------------- 164.1g Ca(NO3)2 ----- 1 molecule1 Ca(NO3)2 - 1 mol NaF= 22.3g CaF236.5g NaF * 1 mol NaF * 1 molecule1 CaF2 * 78.08g CaF2------------ 41.99g NaF --- 2 molecules1 NaF --- 1 mol CaF2= 36.7g CaF2The theoretical yield of CaF2 is 22.3g, as it is the lowest amount of product created. In this case, the limiting reagent (the reactant that produced the least product) is Ca(NO3)2 and the reagent in excess is NaF."A chain is only as strong as its weakest link".
Conversions of stoichiometry?
Stoichiometry Conversions Using Balanced EquationsMol A --> Mol B (Mol -->Mol)("How many moles of B are needed to react with X mol A?") __A + __B --> __AB1. Multiply the number of moles (A) by the number of moles (# molecules)1 for B.2. Divide the result by the number of molecules for A.- FormulaMol A * # molecules B/# molecules A- Conversion factorMol A * # molecules (mol) B = Mol B----- # molecules (mol) A*****1 the number of moles in this case refers to however many molecules of each substance is within the balanced equation (# molecules). For clarity sake I put # molecules in place of "moles" where appropriate; however, on practice problems or demonstrations shown in textbooks, it's likely it will say moles instead of # molecules.*****Mol A --> Mass B (Mol --> Mass)("How many grams B are needed to produce X mol A?") __A + __B --> __AB1. Multiply mol A by # molecules B; divide by # molecules A. (Mol A --> Mol B)2. Multiply result from 1 by molar mass B (mol B --> Mass B).- FormulaMol A x # molecules B/# molecules A- Conversion FactorMol A x # molecules (mol) B x molar mass B = mass B--------- # molecules (mol) A ----- 1 mol BMass A --> Mol B (Mass --> Mol)("How many mol of B are needed to react with X g A?") __A + __B --> __AB1. Multiply Mass A by # molecules B.2. Divide by molar mass multiplied by # molecules A.- FormulaMass A x # molecules B/(Molar Mass A x # molecules A)- Conversion FactorMass A x 1 mol A x # molecules (mol) B = mol B--- molar mass A - # molecules (mol) AMass A --> Mass B (Mass --> Mol --> Mass)("How many grams of A can be produced from X grams B?") __A + __B --> __AB1. Convert from grams (g) to mol for substance A (mass A --> mol A).2. Divide mol A by mol B (mol A --> mol B [Molar Ratio of Substances]).3. Multiply mol B by molar mass B (mol --> mass)In summary, you are converting from grams A to mol A, then mol A to mol B, then mol B to grams B.- FormulaMass A x # molecules B x Molar Mass B/(Molar Mass A x # molecules A)- Conversion FactorMass (g) A * 1 mol A -- x -- # molecules (mol) B -- x -- molar mass B = mass B----- molar mass (g) A --x-- # molecules (mol) A --- x --- 1 mol BLimiting Reagent and Reagent in ExcessThe limiting reagent or limiting reactant is the substance that limits the reaction. ("What substance has the least amount produced from a reaction?"). The reagent in excess or reactant in excess is the product left over or reagent that is leftover from the reaction; in other words, the reagent that has the most product that did not react.To identify the limiting reagent or limiting reactant, identify which substance produced the least amount of product (which reactant yields the least amount of product).Steps (given masses of products):A. Identify amount of product created per reactant (reactant --> product yield).1. Balance the equation if it has not been done already.2. Convert the given masses of reactant (A, B, etc.) to mass product (C) (see "Mass A --> Mass B [Mass --> Mass]" above). (mass reagent --> mass product)B. Identify the limiting reagent. The limiting reagent (reactant) will be the reactant (A, B, etc.) that yields the least amount of C (product).C. Identify the reagent in excess. The reagent in excess (reactant) will be the reactant (A, B, etc.) that yields the most amount of C (product).D. Give how much reagent in excess remain unreacted. (How much reactant is leftover). For simplicity sakes, the limiting reagent will be A and the reagent in excess will be B. "For every X grams of the limiting reagent, there is Y grams of the reagent in excess".Mass Limiting Reactant --> Mass Reagent in Excess:First convert the mass of the limiting reagent to the mass of the reagent in excess (mass limiting reagent : mass reagent in excess [reacting] ratio); then subtract the mass of the limiting reagent from the mass of the reagent in excess (that reacted)1. Convert the mass of the limiting reagent to mass of the reagent in excess (ratio mass limiting reagent: mass reagent in excess). Refer to "Mass A --> Mass B" above.2. Subtract the original amount of B (reagent in excess) from the amount of B needed to react with A (limiting reagent).Reagent in Excess leftover = Starting Mass A - Reacting Mass B (step 1 answer)E. Find % yield.% yield = actual yield (given)------- theoretical yield (must be found)** the theoretical yield is the amount of product theoretically produced by the limiting reagent; the actual yield is the amount of product actually produced by the reactants; the theoretical yield will have been found in step A. The actual yield will be given within the worded problem.Ex. 4Na2CO3 + Fe3Br8 --> 8NaBr + 4CO2 + Fe3O4 A. How many grams of Fe3O4 can be produced from 100.0g Na2CO3 and 300.0g Fe3Br8?100.0g Na2CO3 x 1 mol Na2CO3 x 1 mol Fe3O4 x 231.6g Fe3O4------------------ 106.0g Na2CO3 -- 4 mol Na2CO3 -- 1 mol Fe3O4= 54.62g Fe3O42(2 this is known as the theoretical yield, which will be needed when calculating percentage yield later on).300.0g Fe3Br8 x 1 mol Na2CO3 x 1 molecules Fe3O4 x 231.6g Fe3O4----------------- 106.0g Fe3Br8 ---- 4 molecules Fe3Br8 -- 1 mol Fe3O4= 86.12g Fe3O4B. What is the limiting reagent?The limiting reagent in this case is Na2CO3 because it has the lowest theoretical yield for producing Fe3O4 (54.62g Fe3O4 vs 86.12g Fe3O4).C. What is the reagent in excess?The reagent in excess is Fe3Br8 because it has the highest theoretical yield for producing Fe3O4.D. How many grams of the reagent in excess(Fe3Br8)remain unreacted?100.0g Na2CO3 x 1 mol Na2CO3 x 1 molecules Fe3Br8 x 806.8g Fe3Br8------------------- 106.0g Na2CO3 -- 4 molecules Na2CO3 - 1 mol Fe3Br8= 190.3g Fe3Br8300.0g Fe3Br8 - 190.3g Fe3Br8 = 109.7g Fe3Br8 leftover (unreacted)E. If 42.75g of Fe3O4 were isolated, what is the % yield?% yield = 42.75g----------- 54.62g2 x 100% = 78.27%
Which petroleum refinery product has lowest boiling point?
Gasoline is the petroleum refinery product has lowest boiling point.
What is the lowest multiple of 23 and 9?
Their product.
What are the ingredients in olay face wash?
Olay has more than one face wash product. When looking at the product in the store, each ingredient is listed on the package in order from highest to lowest amount. On their website you can find an ingredient list of each product they currently offer.
How many grams carbon dioxide is made when 42 grams of propane is burned with 115 grams of oxygen how many grams of water is made?
First you need to write a balanced equation. You are given that propane undergoes a combustion reaction that produces carbon dioxide and water.Unbalanced: C3H8 + O2 ---> CO2 + H2OBalanced: C3H8 + 5O2 ---> 3CO2 + 4H2OGivens:42.0 grams C3H8 (Molecular mass 44.0 g)115.0 grams O2 (Molecular mass 32.0 g)Molecular mass of CO2: 44.0 gMole ratio 1:5:3:4 (C3H8:O2:CO2:H2O)Then you need to find which of the reactants are the limiting reactant (lowest value) and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.42.0 g C3H8 / (44.0 g) = .955 moles C3H8 115.0 g
Solids have the lowest amount of what?
Energy
Which is recommended for those products with the lowest volume and the lowest standardization on the product continuum?
Continuous flow
