# What is the theoretical yield of acetylene when 2.38 g of calcium carbide reacts with excess water? What is the percentage yield if I obtained 0.77 g acetylene?

Dec 17, 2016

The theoretical and percentage yields of acetylene are 0.967 g and 80 %, respectively.

#### Explanation:

"M_r:color(white)(ml) 64.10color(white)(mmmmmmmmmmmm) 26.04
$\textcolor{w h i t e}{m m m l} {\text{CaC"_2 + "2H"_2"O" → "Ca(OH)"_2 + "C"_2"H}}_{2}$

For convenience, I have written the molar masses above the formulas.

(a) Calculate the moles of ${\text{CaC}}_{2}$

${\text{Moles of CaC"_2 = 2.38 color(red)(cancel(color(black)("g CaC"_2))) × "1 mol CaC"_2/(64.10 color(red)(cancel(color(black)("g CaC"_2)))) = "0.037 13 mol CaC}}_{2}$

(b) Calculate moles of ${\text{C"_2"H}}_{2}$ formed from the ${\text{CaC}}_{2}$

${\text{0.037 13"color(red)(cancel(color(black)("mol CaC"_2))) × ("1 mol C"_2"H"_2)/(1 color(red)(cancel(color(black)("mol CaC"_2)))) = "0.037 13 mol C"_2"H}}_{2}$

(c) Calculate the theoretical yield of ${\text{C"_2"H}}_{2}$

$\text{Theoretical yield" = "0.037 13" color(red)(cancel(color(black)("mol C"_2"H"_2))) × ("26.04 g C"_2"H"_2)/(1 color(red)(cancel(color(black)("mol C"_2"H"_2)))) = "0.967 g adduct}$

The theoretical yield of product is 0.967 g.

(d) Calculate the percentage yield of ${\text{C"_2"H}}_{2}$

The formula for percentage yield is

color(blue)(bar(ul(|color(white)(a/a)"Percentage yield" = "actual yield"/"theoretical yield" × 100 %color(white)(a/a)|)))" "

"Percentage yield" = ("0.77" color(red)(cancel(color(black)("g"))))/(0.967 color(red)(cancel(color(black)("g")))) × 100 % = 80 %

The percentage yield is 80 %.