Consider this reaction, which occurs in the atmosphere and contributes to photochemical smog: CH4(g) + CCl4(g) →2CH2Cl2(g) If there is 11.45 g CH4 and excess CCl4 present, the reaction yields 94.1 g CH2Cl2. Calculate the percent yield for the reaction.

The percent yield of the reaction is 77.6 %

Explanation:

Step 1: Data given

Mass of CH4 = 11.45 grams

Molar mass of CH4 = 16.04 g/mol

Mass of CH2Cl2 produced = 94.1 grams

Step 2: The balanced equation

CH4(g) + CCl4(g) →2CH2Cl2(g)

Step 3: Calculate moles CH4

Moles CH4 = mass CH4 / molar mass CH4

Moles CH4 = 11.45 grams / 16.04 g/mol

Moles CH4 =  0.714 moles

Step 4: Calculate moles CH2Cl2

For 1 mol CH4 we need 1 mol CCl4 to produce

For 0.714 moles CH4 we'll have 2*0.714 = 1.428 moles CH2Cl2

Step 5: Calculate mass CH2Cl2

Mass CH2Cl2 = moles CH2Cl2 * molar mass CH2Cl2

Mass CH2Cl2 = 1.428 moles * 84.93 g/mol

Mass CH2Cl2 = 121.3 grams

Step 6: Calculate percent yield

% yield = (actual yield / theoretical yield ) * 100%

% yield = (94.1 grams / 121.3 grams ) * 100 %

% yield = 77.6 %

The percent yield of the reaction is 77.6 %

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