stoichiometry help!?

Question

tungsten can be produced from its oxide by reaction the oxide with hydrogen at a high temperature according to the following equation.

WO3 + 3H2 ---> W + 3H2O

a)What is the percent yield if 56.9g of WO3 yields 41.4g of tungsnten?

b)How many moles of tungsten will be produced from 3.72f of WO3 if the yield is 92.0%?

c) a chemist carries out this reaction and obtains 11.4g of tungsten. If the percent yield is 89.4% what mass of WO3 was used?

Answer 1

a)
Theoretical yield = 56.9 g WO3 * 1 mole / 231.85 g * 1 mole W/ 1 mole WO3 * 183.85 g W/mole = 45.1 g W

% yield = actual/theoretical * 100 = 41.4/45.1 * 100 = 91.8%

b)
Theoretical yield = 3.72 g WO3 * 1 mole / 231.85 g * 1 mole W/ 1 mole WO3 = 0.016 moles W

Actual yield = % yield * theoretical yield = 0.92 * 0.016 = 0.0147 moles

c)
Theoretical yield = actual yield / % yield = 11.4 g/0.894 = 12.75 g

12.75 g W * 1 mole/ 183.85 g * 1 mole WO3/1 mole W * 231.85 g WO3/mole = 16.07 g

Answer 2

1. percent yield
step 1: find limiting reactant
= find out how much W would be formed by both reactants.

(gWO3)*(molWO3/gWO3)*molar ratio*(gW/molW)
(gH2)*(molH2/gH2)*molar ratio*(gW/molW)

The one that would yield the least amount of product is your limiting reactant (but you're not giving the amount of H2 used so i'm assuming it was used in excess, which means WO3 is your limiting reactant).

step 2: theoretical yield
equal to amount of product that should have been formed using limiting reactant.

step 3: calculate
% yield = (observed/theoretical)*100

2. this is like algebra,
use the theoretical yield from 3.72 g
then observed yield = X;
for 92.0 = 100*(x/theoretical)
then convert grams to moles

3. this is like algebra also
89.4 = 100*(11.4g/X), solve for x, which equals theoretical
and the rest is like in finding the limiting reactant step:
theoretical(g) *(molW/gW)*molar ratio * (gWO3/molWO3)=