Chapter 12.2, Problem 13SP

Interpretation Introduction

Interpretation: The grams of acetylene are to be stated.

Concept Introduction: According to the law of conservation of mass, the mass can neither be created nor be destroyed. Every reaction must be balanced. With the help of the balanced equation, the limiting reagent can be found and in accordance with it, the amount of product can be calculated.

Answer to Problem 13SP

The grams of acetylene is 2.03 g.

Explanation of Solution

The reaction is given as follows:

  ${\text{CaC}}_{\text{2}}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\left(g\right)+\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right)$

The mol of ${\text{CaC}}_{\text{2}}$ is calculated as:

  $\begin{array}{c}\text{Mole}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{5.00\text{g}}{64.01\text{g/mol}}\\ =0.078\text{mol}\end{array}$

According to the balanced equation,

  $\begin{array}{c}1\text{mol}\text{of}{\text{CaC}}_{\text{2}}\to 1\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\\ 0.078\text{mol}\text{of}{\text{CaC}}_{\text{2}}\to 0.078\text{mol}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\end{array}$

The mass of acetylene is calculated as:

  $\begin{array}{c}\text{Mass}=\text{Mole}\times \text{Molar}\text{mass}\\ =0.078\text{mol}\times 26\text{g/mol}\\ =\text{2}\text{.03}\text{g}\end{array}$

Hence, the grams of acetylene are 2.03 g.