Areaction between methane and hydrogen sulfide to produce carbon disulfide and hydrogen gas was carried out.

a 32.0 g quantity of methane produced 70.0 g of carbon disulfide. what was the percent yield for this reaction?

The percent yield for this reaction was 45.98 %

Explanation:

Let's take a look to the reaction:

CH₄(g)  + 2H₂S (g) → CS₂ (g)  +  4H₂ (g)

We can not apply the limiting reactant's concept, as we don't have any information of H₂S. So let's work only with methane.

To know the moles we must do mass / molar mass

32 g / 16 g/m = 2 moles

Ratio is 1:1, so 2 moles of methane produce 2 moles of disulfide.

Molar mass of CS₂ = 76.12 g/m

Moles . molar mass = 2 m . 76.12 g/m = 152.24 g

This are the moles of gas, with the 100 % yield reaction, we only made 70 g so let's find out the yield percent, by a rule of three.

152.24 g 100 %

70 g (70 . 100) / 45.98 %