Question

An industrial scale process for making acetylene consists of the following sequence of operations:

$\begin{array}{l}CaC{O}_3\to CaO+C{O}_2\\ CaO+3C\to Ca{C}_2+CO\\ Ca{C}_2+2H_{2}O\to Ca{\left(OH\right)}_2+C_2{H}_2\end{array}$

What id the percentage yield of the overall process if 2.32 metric tons is produced from 10.0 metric tons ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{2}}$limestone ( 1 Metric ton = 1000kg )

Short answer

Percentage yield of the overall process is 89.2%

Step-by-step solution

Total Acetylene Produced

The reaction in an industrial process of making the acetylene are

$\begin{array}{l}CaC{O}_3\to CaO+C{O}_2\\ CaO+3C\to Ca{C}_2+CO\\ Ca{C}_2+2H_{2}O\to Ca{\left(OH\right)}_2+C_2{H}_2\end{array}$

According to balanced equation 1 mol $\mathrm{CaCO}3$will produce 1mol${\mathrm{c}}_{2\mathrm{H}2}$

Then amount of ${\mathrm{C}}_{2\mathrm{H}2}$produced from 10.0 metric tons of ${\mathrm{Caco}}_3$IS:

$\begin{array}{l}\left(10.0MetricTon\right)\times \frac{{10}^{6}g}{IMT}\times \frac{1molCaC{O}_3}{100.09gCaC{O}_3}\times \frac{1mol{C}_2{H}_2}{1molCaC{O}_3}\times \frac{26.04g{C}_2{H}_2}{1mol{C}_2{H}_2}\\ =2.60\times {10}^{6}g{C}_2{H}_2\times \frac{1kg}{{10}^{3}kg}\times \frac{1MetricTon}{1000kg}\\ =2.60MetricTons\end{array}$

Therefore yield of acetylene is 2.60 metric tonstheoretically

Percentage Yield

$\begin{array}{c}PercentageYield=\frac{ActualYield}{TheoriticalYield}\times 100\\ =\frac{2.32g}{2.60g}\times 100\%=89.2\%\end{array}$

Hence the percentage yield is 89.2%