Chapter 9, Problem 58A

Interpretation Introduction

Interpretation: The mass of oxygen needed for complete combustion of 150 g of acetylene needs to be calculated.

Concept introduction: The ratio of mass of substance to its molar mass is said to be number of moles of that substance.

Answer to Problem 58A

The mass of oxygen needed for complete combustion of 150 g of acetylene is 460.77 g.

Explanation of Solution

The balanced complete chemical equation for the combustion of acetylene is:

  ${\text{2C}}_{\text{2}}{\text{H}}_{\text{2}}\left(\text{g}\right){\text{+ 5O}}_{\text{2}}\left(\text{g}\right)\to {\text{4CO}}_{\text{2}}\left(\text{g}\right){\text{+ 2H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The molar mass of acetylene is 26.04 g/mol. The number of moles of acetylene is calculated as shown:

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Number of moles}=\frac{\text{150 g}}{\text{26}\text{.04 g/mol}}\\ \text{Number of moles}=5.76\text{ mol}\end{array}$

From the balanced chemical reaction, the mole ratio of acetylene and oxygen is 2:5 so, the number of moles of oxygen for 5.76 mol of acetylene is:

  $\text{5}{\text{.76 mol of C}}_{\text{2}}{\text{H}}_{\text{2}}\text{×}\frac{{\text{5 mol of O}}_{\text{2}}}{{\text{2 mol of C}}_{\text{2}}{\text{H}}_{\text{2}}}\text{ = 14}{\text{.4 mol of O}}_{\text{2}}$

Now, the mass of oxygen needed for complete combustion of 150 g of acetylene is calculated as:

The molar mass of oxygen is 31.998 g/mol. So,

  $\begin{array}{l}\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}\\ \text{14}\text{.4 mol}=\frac{\text{Mass}}{\text{31}\text{.998 g/mol}}\\ \text{Mass = 14}\text{.4 mol}\times \text{31}\text{.998 g/mol}\\ \text{Mass = }460.77\text{ g}\end{array}$

Hence, the mass of oxygen needed for complete combustion of 150 g of acetylene is 460.77 g.

Conclusion

460.77 g is the mass of oxygen needed for complete combustion of 150 g of acetylene.