Methane and chlorine react to form chloromethane, CH3Cl and hydrogen chloride. When 29.8 g of methane and 40.3 g of chlorine gas undergo a reaction that has a 64.7% yield, how many grams of chloromethane form

18.26g

Explanation:

Equation of reaction;

CH₄ + Cl₂ → HCL + CH₃Cl

mass of CH₄ = 29.8g

molar mass of CH₄ = 16g/mol

mass of Cl₂ = 40g

molar mass of Cl₂ = (35.5*2) = 71g/mol

molar mass of CH₃Cl = 50.46g/mol

no. of moles = mass / molar mass

no. of moles of CH₄ = 29.8 / 16 = 1.8625 moles

no. of moles of Cl₂ = 40 / 71 = 0.56moles

Note : limiting reactant is Cl₂

From equation of reaction,

1 mole of Cl₂ = 1 mole of CH₃Cl

0.56 moles of Cl₂ = 0.56 moles of CH₃Cl

No. Of moles of experimental yield = no. Of moles of theoretical yield * percentage yield.

Theoretical yield of n(CH₃Cl) = 0.56 moles

n(CH₃Cl exp) = 0.56 * 0.647

n(CH₃Cl exp) = 0.36232 moles

Number of moles = mass / molar mass

Mass = no. Of moles * molar mass

Mass = 0.36232 * 50.46 = 18.26g

m(CH₃Cl) = 18.26g

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