Question

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


`CH_4(g) + 2H_2S(g) \rightarrow CS_2 (g) + 4H_2(g)`
A 32.0 g quantity of methane produced 70.0 g of carbon disulfide. What was the percent yield for this reaction?

Answer 1

The percent yield for this reaction was 45.98 %

Step-by-step 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 %