Question

Acetylene is synthesized using 4.56 g of calcium carbide and enough water to complete the reaction. The residue left behind was measured to be 9.45 mL an aqueous Ca(OH)2. The residue was titrated with 0.500 M HCl and endpoint was reached after 2.55 mL was added. What is the percent yield of the reaction?

Acetylene – 26.04 g/mol; calcium carbide – 64.099 g/mol; calcium hydroxide – 74.093 g/mol

PLEASE LIMIT YOUR ANSWER TO 4 S.F.

Answer 1

$$CaC_2 + 2H_2O \rightarrow Ca(OH)_2 + C_2H_2$$

Find theoretical mass of Ca(OH)2:

$n(CaC_2) = \frac{m}{M} = \frac{4.56}{64.099}=0.07114 mol$

$n(Ca(OH)_2) = n(C_2H_2) = 0.07114 mol$

$m_{theor}(Ca(OH)_2) = n\times M = 0.0714 \times 74.093 = 5.271 g$

Find actual mass of Ca(OH)2:

$$Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O$$

$n(HCl )= c\times V = 0.500 \times 0.00255 = 0.001275 mol$

$n(Ca(OH)_2) = \frac{n(HCl)}{2} = \frac{0.001275}{2} = 0.0006375 mol$

$m_{actual} (Ca(OH)_2) = n\times M = 0.0006375 \times 74.093 = 0.04723 g$

$$Yield = \frac{m_{actual}}{m_{theoretical}}\times 100 \% =\frac{0.04723}{5.271}\times 100\% = 0.8961 \%$$