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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

Respuesta :

aabdulsamir

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

pstnonsonjoku

The actual yield of chloromethane is 18.6 g of chloromethane.

The equation of the reaction is;

CH4 + Cl2 -----> CH3Cl + HCl

Number of moles of methane =  29.8 g/ 16 g/mol = 1.86 moles

Number of moles of chlorine = 40.3 g/71 g/mol = 0.57 mole

From the reaction equation, we can see that chlorine is the limiting reactant hence the yield of chloromethane depends on the number of mole of chlorine gas.

Hence, theoretical yield of chloromethane =  0.57 mole × 50.49 g/mol

= 28.77 g

We know that percent yield = actual yield /theoretical yield × 100/1

Actual yield of chloromethane =  64.7  × 28.77 g/100

= 18.6 g of chloromethane

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