Question

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

Answer 1

Answer:

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

Answer 2

Answer:

37.091g

Explanation:

The reaction is given as;

Methane + Chlorine --> Chloromethane

CH4 + Cl2 --> CH3Cl + HCl

From the reaction above;

1 mol of methane reacts with 1 mol of chloromethane.

To proceed, we have to obtain the limiting reagent,

28.9g of methane;

Number of moles = Mass / molar mass = 29.8 / 16 = 1.8625 mol

40.3g of chlorine;

Number of moles = Mass / molar mass = 40.3 / 35.5= 1.1352 mol

Since the reaction has a stoichiometry ratio of 1:1, the limiting reagent is the chlorine

From the reaction,

1 mol of chlorine yirlds 1 mol of chloromethane

1.1352 yields x?

1 = 1

1.1352 = x

x = 1.1352

Mass = Number of moles * Molar mass = 1.1352 * 50.5 = 57.3276

Since the reaction has a 64.7% yield, mass of chloromethane formed is given as;

0.647 * 57.3276 = 37.091g