Question

Acetylene gas, C2H2, can be produced by the reaction of calcium carbide and water. CaC2(s) + 2H2O(l) --> C2H2(g) + Ca(OH)2(aq) How many liters of acetylene at 742 mm Hg and 26C can be produced from 2.54 g CaC2(s)?

Answer 1

Answer:

1.0 L

Explanation:

Given that:-

Mass of $CaC_2$ = $2.54\ g$

Molar mass of $CaC_2$ = 64.099 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{2.54\ g}{64.099\ g/mol}$

$Moles_{CaC_2}= 0.0396\ mol$

According to the given reaction:-

$CaC_2_{(s)} + 2H_2O_{(l)}\rightarrow C_2H_2_{(g)} + Ca(OH)_2_{(aq)}$

1 mole of $CaC_2$ on reaction forms 1 mole of $C_2H_2$

0.0396 mole of $CaC_2$ on reaction forms 0.0396 mole of $C_2H_2$

Moles of $C_2H_2$ = 0.0396 moles

Considering ideal gas equation as:-

$PV=nRT$

where,

P = pressure of the gas = 742 mmHg  

V = Volume of the gas = ?

T = Temperature of the gas = $26^oC=[26+273]K=299K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles = 0.0396 moles

Putting values in above equation, we get:

$742mmHg\times V=0.0396 mole\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\V=\frac{0.0396\times 62.3637\times 299}{742}\ L=1.0\ L$

1.0 L of acetylene  can be produced from 2.54 g $CaC_2$.