Question

What is the percent yield of a reaction in which 41.6 g of tungsten (VI) oxide (WO3) reacts with excess hydrogen gas to produce metallic tungsten and 9.44 mL of water (d = 1.00 g/mL)?

Answer 1

to calculate the percent yield you need the actual yield and the theoretical yield,

\(\%~yield=\dfrac{actual~yield}{theoretical~yield}*100\%\)

Answer 2

ok but how do i get actual and theoretical yield?

Answer 3

the actual yield is given in terms of the amount of water produced.
the theoretical yield is found by using the mass of WO3 and the stoichiometric ratio (i.e. like most other stoichiometry problems).

Answer 4

so the acutual is 9.44?

but can you help me set up the stoich thing up and stuff step by step?

Answer 5

yep, make sure you find the theoretical for the amount of water produced, since you're using the mass of water.

For the theoretical yield first convert the mass of reactant (i.e. WO3) to moles, then use the stoichiometric coefficients to find moles of water produced. Set up a ratio using the species of interest, like so:

e.g. for a general reaction:

\(\color{red}{a}A + \color{blue}{b}B\) \(\rightleftharpoons\) \( \color{green}{c}C\)

where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients ,

\(\dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\)

From here you can isolate what you need.
For example: if you have 2 moles of B, how many moles of C can you produce?

solve algebraically: \(\dfrac{2}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\rightarrow n_C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\)
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To convert mass to moles, use the relationship:

\(n=\dfrac{m}{M}\)

where, M=molar mass, m=mass, and n= moles.

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After you've found this, use the formula i posted in my first reply to find the percent yield.

Answer 6

thanks

Answer 7

no probs