Question

# When 25.6 g of methane and 40.6 g of chlorine gas undergo a reaction that has...

When 25.6 g of methane and 40.6 g of chlorine gas undergo a reaction that has a 56.2% yield, what mass of chloromethane (CH3Cl) forms? Hydrogen chloride also forms.

The reaction is,

CH4 + Cl2 ----> CH3Cl + HCl

Given,

Mass of CH4 = 25.6 g

Mass of Cl2 = 40.6 g

We know that,

Molar Mass of CH4 = 16 g / mol

Molar Mass of Cl2 = 71 g / mol

=> Moles of CH4 = 25.6 / 16 = 1.6 moles

Moles of Cl2 = 40.6 / 71 = 0.572 moles

According to the stoichiometry of the reaction 1 mole of CH4 reacts with 1 mole of Cl2 to prpoduce 1 mole of CH3Cl and 1 mole of HCl.

We see that we have 1.6 moles of CH4 but only 0.572 moles of Cl2. Hence Cl2 will be the limiting reagent.

Therefore, 0.572 moles of Cl2 should produce 0.572 moles of CH3Cl (theoretically).

But the yield of the reaction is 56.2 %

=> Moles of CH3Cl produced = 0.572 x 0.562 = 0.32 moles

Molar Mass of CH3Cl = 50.5 g / mol

=> Mass of CH3Cl produced = 0.32 x 50.5 = 16.16 g

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