Question

When 37.1 g of CH4 reacts completely with excess chlorine yielding 52.0 g of HCl, what is the percentage yield, according to CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) Group of answer choices

Answer 1

Percent yield = 63.7%

Step-by-step 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%