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?

Question

The equation is #4H_2(g) + CS_2(g) -> CH_4(g) + 2H_2S(g)#.

Noah Aldrich · Accepted Answer

#97.5%#

Two things are necessary to know in order to calculate the percent yield of a reaction.

In your case, the problem provides the actual yield. You know that once the reaction is complete, you are left with #"69.8 g"# of methane, #"CH"_4#.

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

The chemical equation in balance

#color(blue)(4)"H"_text(2(g]) + CS_text(2(g]) -> CH_text(4(g]) + 2H_2S_text((g])#.

tells you that every mole of carbon sulfide requires #color(blue)(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 the molar mass of hydrogen gas to convert 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#

In theory, this many moles of hydrogen gas would yield

#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#

Determine how many grams would contain this many moles using the molar mass of methane.

#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 #"71.6 g"# of methane, which implies that its percent yield was smaller than #100%#.

Use the following formula to find the reaction's percent yield:

#color(blue)("% yield" = "what's actually produced"/"what is theoretically produced" xx 100)#

In your situation, you'll have

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

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

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

Luna Chen · Answer

undefined Percent yield is calculated using the formula:

$$ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 $$

First, find the theoretical yield of CH₄:

$$ \text{Molar mass of H₂} = 2 \times 1 = 2 \, \text{g/mol} $$
$$ \text{Molar mass of CH₄} = 12 + 4 = 16 \, \text{g/mol} $$

$$ \text{Moles of H₂} = \frac{36 \, \text{g}}{2 \, \text{g/mol}} = 18 \, \text{mol} $$

$$ \text{Theoretical moles of CH₄} = \frac{18 \, \text{mol}}{3} = 6 \, \text{mol} $$

$$ \text{Theoretical yield of CH₄} = 6 \, \text{mol} \times 16 \, \text{g/mol} = 96 \, \text{g} $$

Now, use the formula for percent yield:

$$ \text{Percent Yield} = \left( \frac{69.8 \, \text{g}}{96 \, \text{g}} \right) \times 100 $$

$$ \text{Percent Yield} = 72.92\% $$