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

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%

explanation: no but i do not usually comment

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer