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A plus 2B yields 3C If he starts with 5 grams of A and 8 grams of B is B the limiting reactant and why?
Wiki User
∙ 14y ago
A + 2B = 3C
5A + 10B = 15C
4A + 8B = 12C
A and B are the reactants. As there is not enough B to use all the A, B is the limiting factor in the production of C.
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∙ 14y ago
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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
If there are 4 grams of reactant how many grams of product are produced by the chemical reaction?
4
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%
In the explosion of a hydrogen filled balloon 0.90 g of hydrogen reacted with 7.2 g of oxygen to form how many grams of water vapor?
The stoichiometric equation is 2H2 + O2 --> 2H2O. From the given amounts, there are .645 moles of H2 and .325 moles of O2. The limiting reactant is H2, so .645 moles of H2O are formed which is equal to 11.6 grams.
If .3 mole of N2 are reacted with .6 mole of H2 what mass of NH3 could form the formula N2 plus 3 H2 makes 2 NH3 The answer to the problem is 6.8 but don't know how to get that?
First you have to find the limiting reactant. You have .3 moles of nitrogen and .6 moles of hydrogen, but you don't know which one is going to run out first.In any of these stoichiometry problems, you need to write down the formula:N2 + 3H2 → 2NH3Take both nitrogen and hydrogen and figure out how much ammonia is made alone..6 moles Hydrogen ÷ 3 moles hydrogen × 2 moles ammonia = .4 moles ammonia made.3 moles Nitrogen ÷ 1 mole nitrogen × 2 mole ammonia = .6 moles ammonia madeNow you figured out that hydrogen is the limiting reactant and the nitrogen is the excess because less ammonia is made using hydrogen. This measurement is what you will be using for the rest of the problem.Take the limiting reactant and use stoichiometry to find how much ammonia can be made.You could start with .6 moles of hydrogen and do the same conversion as above, but add the step of converting to grams. Or, since you already found out that .4 moles ammonia is made, just convert it to grams. The molecular mass of ammonia is 17.0 grams..4 moles ammonia × 17.0 grams = 6.8 grams ammonia
What is the limiting reactant when 41.4 grams of cuo are exposed to 2.56 grams of h2 according to the equation cuo h2 yields cu h2o?
cuo
87.2 grams of glucose burned in 87.2 grams oxygen what is limiting reactant?
Glucose
What is the limiting reactant when 3.14g of nitrogen react with 2.79g of hydrogen to produce ammonia and how many grams of ammonia are produce?
Nitrogen is the limiting reactant and 4.15g of ammonia are produced.
If 25.0 grams of potassium chlorate decomposes, how many grams of potassium chloride will be produced What is the limiting reactant?
I Don't knows Sorry
How many moles of which reactant will remain if 1.39 moles of N and 3.44 moles of H will react to form ammonia find out how many grams of ammonia can be formed and how many moles of limiting reactant?
3,44 moles H2 react with 1,146 moles NH3. The limiting reactant is hydrogen. O,244 moles N2 remain. 19,5 g NH3 are obtained.
In a chemical reaction 300 grams of reactant A are combined with 100 grams of reactant B . both A and B reactant to completion. how much will the product weigh?
200
What is the limiting reactant of 6.5g of C3H8O and 12.3g of O2?
The reaction between Isopropyl alcohol and oxygen is 2 C3H8O + 9 O2 equals 6 CO2 + 8 H2O. So for every mole of isopropyl alcohol, 4.5 moles of oxygen are consumed. 6.5 grams of C3H8O is .108 moles and 12.3 grams of O2 is .384 moles. This means that O2 is the limiting reactant as it needs .486 moles of O2 to finish.
When 10.0 grams of calcium reacts with 20.0 grams of chlorine gas, how many grams of calcium chloride can be produced Which reactant is in excess and which is the limiting reactant?
The balanced chemical equation for the reaction between calcium and chlorine gas to produce calcium chloride is: Ca + Cl2 -> CaCl2. From this equation, we can see that one mole of calcium reacts with one mole of chlorine gas to produce one mole of calcium chloride. The molar mass of calcium is 40.08 g/mol and the molar mass of chlorine gas is 70.90 g/mol. This means that 10.0 grams of calcium is equivalent to 0.249 moles of calcium and 20.0 grams of chlorine gas is equivalent to 0.282 moles of chlorine gas. Since the ratio of calcium to chlorine gas in the balanced chemical equation is 1:1, this means that 0.249 moles of calcium would react completely with 0.249 moles of chlorine gas, leaving an excess of 0.033 moles (or 2.34 grams) of chlorine gas. The limiting reactant in this reaction is calcium, and the maximum amount of calcium chloride that can be produced is equivalent to the number of moles of the limiting reactant, which is 0.249 moles (or 27.8 grams) of calcium chloride.
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 reactant of 10.0 grams of total mass What is the chemical reaction?
It is possible to have ten grams of anything. Specifying the total weight tells you nothing about which reactant you have or what reaction it will undergo.
If there are 4 grams of reactant how many grams of product are produced by the chemical reaction?
4
How do you concert grams to kilograms?
1000 grams = 1 kilogram So dividing grams by 1000 yields kilograms
