Chemistry help: Chemical Reactions and Stoichiometry, [Percent Yield]?

Question

1. Tungsten (W) can be produced from its oxide by reacting the oxide with hydrogen at high temperatures, according to the following reaction:
WO3 + 3H2 → W + 3H2O

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

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

c) This reaction is carried out in the lab, and 11.4 g of tungsten are obtained. If the percent yield is 89.4%, what mass of WO3 was used?

Answer 1

The stoichiometry of this problem: 1 mol of WO3 yields 1 mol W. This 1:1 ratio should simplify the problem a lot for you.

The atomic weight of W is 183.84 g/mol.

The atomic weight of O is 15.9994 g/mol. The molecular weight of WO3 is 183.84 + 3*15.9994 = 231.8382 g/mol.

(a) We just need to find the number of moles of W produced, divide by the number of moles of WO3 at the start, and multiply by 100%.

41.4 g / 183.84 g/mol = 0.225196 mol W
56.9 g / 231.8382 g/mol = 0.24543 mol WO3

(0.225196 mol W/0.24543 mol WO3)*(1 mol WO3/1 mol W)*100% = 91.8%
--> Note: the second quantity containing the 1:1 stoichiometry for use in problems when stoichiometry is different.

(b) First, determine the number of moles of WO3:

3.72 g / 231.8382 g/mole = 0.016046 mol WO3

Now we can determine the number of moles of W produced based on the give percent yield. Start with the simple equation, plug in the known values, then solve the equation for # moles W:

(# moles W/# moles WO3)*(1 mol WO3/1 mol W)*100%=percent yield

(# moles W/0.016046 mol WO3)*(1 mol WO3/1 mol W)*100%=92.0%

# moles W = 0.016046*92.0/100 = 0.01476 mol W

(c) Remember, percent yield is based on # moles NOT number of grams. So, let's start by converting the mass of W to moles of W.

11.4 g / 231.8382 g/mol = 0.082109 mol W

Again we'll use with the equation:

(# moles W/# moles WO3)*(1 mol WO3/1 mol W)*100%=percent yield

Now let's plug in the known values:

(0.082109 mol W/# moles WO3)*(1 mol WO3/1 mol W)*100% = 89.4%

Rearrange to solve for # moles WO3:

# moles WO3 = 0.082109*(100/89.4) = 0.091844 mol WO3

All that's left is to convert # moles WO3 to # grams WO3 by using the molecular weight:

0.091844 * 231.8382 = 21.3 g WO3

--> Note: all answers should contain three significant digits based on the wording of the problems.

Answer 2

a) The theoretical yield is one mole of W per mole of WO3

mm(WO3) = 231.85, so 56.9/231.85 = 0.2454moles

mm(W) = 183.84, so moles of W produced is 41.4/183.84 = 0.2252 moles

The yield is then 0.2252/0.2454 = 91.8%

b) 3.72g of WO3 = 3.72/231.85 = 0.016 moles
At 92% yield, this produces 0.0148 moles of W, which is
0.0148*183.84 = 2.71g

c) 11.4g of W is 11.4/183.84 = 0.062 moles. For 89.4% yield, that means 0.062/0.894 = 0.0694 moles of CO3 were used. That represents 0.0694*231.85 = 16.1g CO3

Answer 3

Use the Mole Concept.