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When 37.1 g of CH₄ reacts completely with excess chlorine, yielding 52.0 g of HCl, what is the percentage yield according to the reaction:

\[ \text{CH}_4(g) + \text{Cl}_2(g) \rightarrow \text{CH}_3\text{Cl}(g) + \text{HCl}(g) \]

Answer :

Answer:

Percent yield = 63.7%

Explanation:

Given data:

Mass of CH₄ = 37.1 g

Actual yield of HCl = 52.0 g

Percent yield = ?

Solution:

Chemical equation:

CH₄ + Cl₂ → CH₃Cl + HCl

Number of moles of CH₄:

Number of moles = mass/ molar mass

Number of moles = 37.1 g/ 16 g/mol

Number of moles = 2.3 mol

Now we compare the moles of HCl with CH₄.

CH₄ : HCl

1 : 1

2.3 : 2.3

Theoretical yield:

Mass of HCl = number of moles × molar mass

Mass of HCl = 2.3 mol × 35.5 g/mol

Mass of HCl = 81.65 g

Percent yield:

Percent yield = Actual yield / theoretical yield × 100

Percent yield = 52.0 g/ 81.65 g × 100

Percent yield = 63.7%

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