# Methane and hydrogen sulfide form when 36g H_2 reacts with carbon disulfide. What is the percent yield if the actual yield of CH_4 is 69.8g?

## The equation is $4 {H}_{2} \left(g\right) + C {S}_{2} \left(g\right) \to C {H}_{4} \left(g\right) + 2 {H}_{2} S \left(g\right)$.

Feb 14, 2016

97.5%

#### Explanation:

In order to determine a reaction's percent yield, you need to know two things

• the actual yield of the reaction
• the theoretical yield of the reaction

In your case, the problem provides the actual yield. You know that once the reaction is complete, you are left with $\text{69.8 g}$ of methane, ${\text{CH}}_{4}$.

Now, the theoretical yield of the reaction is what you would see produced at a 100% yield.

The balanced chemical equation

$\textcolor{b l u e}{4} {\text{H}}_{\textrm{2 \left(g\right]}} + C {S}_{\textrm{2 \left(g\right]}} \to C {H}_{\textrm{4 \left(g\right]}} + 2 {H}_{2} {S}_{\textrm{\left(g\right]}}$.

tells you that every mole of carbon sulfide requires $\textcolor{b l u e}{4}$ moles of hydrogen gas and produces one mole of methane.

Since no information was provided about the mass of carbon disulfide, you can assume that it is in excess. So, if the reaction had a 100%, what mass of methane would you expect that much hydrogen gas to produce?

Use hydrogen gas' molar mass to convert the grams to moles

36 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016 color(red)(cancel(color(black)("g")))) = "17.857 moles H"_2

This many moles of hydrogen gas would theoretically produce

17.857color(red)(cancel(color(black)("moles H"_2))) * "1 mole CH"_4/(color(blue)(4)color(red)(cancel(color(black)("moles H"_2)))) = "4.464 moles CH"_4

Use methane's molar mass to find how many grams would contain this many moles

4.464 color(red)(cancel(color(black)("moles CH"_4))) * "16.04 g"/(1color(red)(cancel(color(black)("mole CH"_4)))) = "71.6 g"

Now, you know that your reaction produced less than $\text{71.6 g}$ of methane, which implies that its percent yield was smaller than 100%.

To get the reaction's percent yield, use the formula

$\textcolor{b l u e}{\text{% yield" = "what's actually produced"/"what is theoretically produced} \times 100}$

In your case, you will have

"% yield" = (69.8 color(red)(cancel(color(black)("g"))))/(71.6color(red)(cancel(color(black)("g")))) xx 100 = 97.49%

You should round this off to two sig figs, since that's how many sig figs you have for the mass of hydrogen gas, but I'll leave it rounded to three sig figs

"% yield" = color(green)(97.5%)