Chapter 12.2, Problem 23LC

(a)

Interpretation Introduction

Interpretation:

The mass of ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ that are produced when $\text{ 52}{\text{.0 g C}}_2{\text{H}}_2$ burn in oxygen needs to be calculated.

Concept Introduction :

The balanced equation is the one that has the same number of atoms of each species in both the reactant and product side. Mole conversion factor is used to do further stoichiometric calculations using a balanced equation.

Answer to Problem 23LC

The mass of ${\text{CO}}_2$ and ${\text{H}}_2\text{O}$ that are produced are $\text{175}{\text{.15 g CO}}_2$ and $\text{35}{\text{.83 g H}}_2\text{O}$ .

Explanation of Solution

The combustion reaction of acetylene is:

  ${\text{2 C}}_2{\text{H}}_2(g)+5O_2(g)\to 4{\text{CO}}_2\text{(}g{\text{)+ 2H}}_2\text{O(}g\text{)}$

Step 1: Convert the given quantity to moles.

  $\begin{array}{l}\text{No}\text{. of moles = }\frac{\text{Given mass}}{\text{Molar mass}}\\ \text{ =}\frac{52.0g}{26.04\frac{g}{mol}}\\ \text{ = 1}\text{.99 moles}\end{array}$

Step 2: Use the mole ratio conversion to determine the moles of the desired substance.

  $\frac{{\text{1 mol CO}}_{\text{2}}}{\frac{\text{1}}{\text{2}}{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{2}}}\text{×1}{\text{.99 mol C}}_{\text{2}}{\text{H}}_{\text{2}}\text{=3}{\text{.98 mol CO}}_{\text{2}}$

The mass of carbon dioxide produced:

  $\text{3}{\text{.98 mol CO}}_{\text{2}}\text{×44}\text{.01}\frac{\text{g}}{\text{mol}}\text{=175}\text{.15 g}$

The moles of water produced:

  $\frac{{\text{1 mol H}}_2\text{O}}{{\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{2}}}\text{×1}{\text{.99 mol C}}_{\text{2}}{\text{H}}_{\text{2}}\text{=1}{\text{.99 mol H}}_2\text{O}$

The mass of water produced:

  $\text{1}{\text{.99 mol H}}_{\text{2}}\text{O × 18}\text{.01}\frac{\text{g}}{\text{mol}}\text{ = 35}\text{.83 g}$

(b)

Interpretation Introduction

Interpretation:

The number of moles of water produced when $\text{64}{\text{.0 g C}}_{\text{2}}{\text{H}}_{\text{2}}$ burns need to be determined.

Concept Introduction :

Combustion reaction is a type of reaction in which a substance burns in the presence of oxygen to form water and carbon dioxide.

Answer to Problem 23LC

The mole of water produced when $\text{64}{\text{.0 g C}}_{\text{2}}{\text{H}}_{\text{2}}$ burns is 2.45 mol.

Explanation of Solution

The combustion reaction of acetylene is:

  ${\text{2 C}}_2{\text{H}}_2(g)+5O_2(g)\to 4{\text{CO}}_2\text{(}g{\text{)+ 2H}}_2\text{O(}g\text{)}$

  $\begin{array}{l}\text{No}\text{. of moles = }\frac{\text{Given mass}}{\text{Molar mass}}\\ \text{ = }\frac{64.0g}{26.04\frac{g}{mol}}\\ \text{ = 2}\text{.45 moles}\end{array}$

The mole ratio conversion to determine the moles of the desired substance.

  $\frac{{\text{1 mol H}}_{\text{2}}\text{O}}{{\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{2}}}\text{×2}{\text{.45 mol C}}_{\text{2}}{\text{H}}_{\text{2}}\text{=2}{\text{.45 mol H}}_{\text{2}}\text{O}$