0
0.5 mol of Na2CO2 is the ACTUAL dry crystal amount.
0.5 M 0f Na2CO3 means that o.5 moles has been dissolved in 1 litre of water.
The M means ' moles per litre(dm^3)'.
lenpollock
Wiki User
∙ 10y ago- difference between 0.50mol na2co3 anf 0.50 M of na2co3
Add your answer:
Earn +20 pts
How many grams are in 0.577 mol Na2CO3?
1 mole Na2CO3 = 105.98844g 0.577mol x 105.98844g/mol = 61.2g Na2CO3
Convert the mass of Na2CO3 to moles with the molecular weight of 105.989?
the answer to your question is 0.0004 g/mol.
A chemist requires 0.277 mol Na2CO3 for a reaction. How many grams does this correspond to?
0.277 mole Na2CO3 x 106 g/mole = 29.4 g (to 3 significant figures)
How many moles of ions are produced when 2 mol of Na2CO3 dissociate?
Na2CO32 * 2 = 4 moles sodium.===========================
How do you prepare 250ml 15m solution of naco3 in a volumetric flask do presume a solvent ie h20 but then you have to presume volumes and conc and because 15M wanted you cant use the flask to dilute?
Did you mean 250 mL of 0.15 M Na2CO3? (It's impossible to make a 15 M Na2CO3 solution, as Na2CO3 is not that soluble.) Yes, you can assume that this will be an aqueous solution. Steps. 1. Calculate the mass of solid Na2CO3 needed. 2. Place this mass of Na2CO3 in the volumetric flask. 3. Add some water and swirl to dissolve the Na2CO3. 4. Carefully add more water until the total volume of solution is 250 mL, as indicated by the line etched on the neck of the volumetric flask. 250 mL x 1 L x 0.15 mol Na2CO3 x 105.99 g Na2CO3 = 4.0 g Na2CO3 needed ........... 1000 mL ......... 1 L ..................1 mol Na2CO3
How many grams are in 0.577 mol Na2CO3?
1 mole Na2CO3 = 105.98844g 0.577mol x 105.98844g/mol = 61.2g Na2CO3
How many moles of washing soda are there in 5g?
Washing soda is sodium carbonate, Na2CO3. Using the atomic weights from the periodic table and the subscripts in the formula, the molar mass of Na2CO3 = 106g/mol. 5g Na2CO3 x (1mol Na2CO3/106g/mol) = 0.05mol Na2CO3
How many grams of ammonia are presented in 5.0 l of a 0.050 m solution?
4.25 grams. .050 M = .050 mol/1 L 5.0 L x .050 mol/L (cancel out L to get mol as a unit)= .25 mol Atomic mass of Ammonia (NH3)= 17 g/mol .25 mol x 17 g/mol (cancel out mol to get g as a unit)= 4.25 g
What is the mass of 7050 mol of uranium?
The mass of 7 050 moles of natural uranium is 238,02891 x 7 050.
How many moles of ions are produced by the dissociation of 0.5 mol of Na2CO3?
1.5 moles
Convert the mass of Na2CO3 to moles with the molecular weight of 105.989?
the answer to your question is 0.0004 g/mol.
A chemist requires 0.277 mol Na2CO3 for a reaction. How many grams does this correspond to?
0.277 mole Na2CO3 x 106 g/mole = 29.4 g (to 3 significant figures)
How many moles of ions are produced when 2 mol of Na2CO3 dissociate?
Na2CO32 * 2 = 4 moles sodium.===========================
How do you prepare 250ml 15m solution of naco3 in a volumetric flask do presume a solvent ie h20 but then you have to presume volumes and conc and because 15M wanted you cant use the flask to dilute?
Did you mean 250 mL of 0.15 M Na2CO3? (It's impossible to make a 15 M Na2CO3 solution, as Na2CO3 is not that soluble.) Yes, you can assume that this will be an aqueous solution. Steps. 1. Calculate the mass of solid Na2CO3 needed. 2. Place this mass of Na2CO3 in the volumetric flask. 3. Add some water and swirl to dissolve the Na2CO3. 4. Carefully add more water until the total volume of solution is 250 mL, as indicated by the line etched on the neck of the volumetric flask. 250 mL x 1 L x 0.15 mol Na2CO3 x 105.99 g Na2CO3 = 4.0 g Na2CO3 needed ........... 1000 mL ......... 1 L ..................1 mol Na2CO3
Volume of 0.2500 M Cobalt III sulfate is required to react completely with 5.00 g of sodium carbonate?
First, write a balanced equation for the information given: Balanced Equation:Co2(S04)3(aq) + 3Na2CO3 (aq)Ã 3Na2SO4 (aq) + Co2(CO3)3 (s)Next, convert the grams of sodium carbonate to moles 5.00 g Na2CO3 x (1molNa2Co3/105.99 g Na2CO3) =0.0472 mol Na2CO3From this, determine the moles of Co2(S04)30.0472 mol Na2CO3 x (1mol Co2(S04)3/3mol Na2CO3) = 0.0157 mol Co2(S04)3Finally, determine volume of Co2(S04)3V = 0.0157mol/0.2500 mol/L = 0.0628 L Co2(S04)3
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 best compares 1 mol of sodium chloride to 1 mol of aluminum chloride?
Each compound has a different molar mass.
