Q. 7.104 - An Introduction to General, Organic, and Biological Chemistry

Question

When nitrogen dioxide (${\mathrm{NO}}_2$) gas from car exhaust combines with water vapor in the air, it forms aqueous nitric acid (${\mathrm{HNO}}_3$), which causes acid rain, and nitrogen oxide gas. ($7.4, 7.7, 7.8$)

a. Write the balanced chemical equation. 

b. How many moles of each product are produced from $0.250$ mole of ${\mathrm{H}}_2\mathrm{O}$?

c. How many grams of ${\mathrm{HNO}}_3$ are produced when $60$ g of ${\mathrm{NO}}_2$ completely reacts?

d. How many grams of ${\mathrm{NO}}_2$ are needed to form $75.0$ g of ${\mathrm{HNO}}_3$?

Step-by-Step Solution

Answer

(a) The required balanced equation is $3{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{HNO}}_3+\mathrm{NO}$

(b) The moles of products produced are style="max-width: none; vertical-align: -4px;" $0.5$ mole of ${\mathrm{HNO}}_3$ and style="max-width: none; vertical-align: -4px;" $0.25$ mole of $\mathrm{NO}$

(c) The $82.34$ g of ${\mathrm{HNO}}_3$ is produced when style="max-width: none; vertical-align: -4px;" $60$ g of ${\mathrm{NO}}_2$ reacts

(d) The $54.78$ g of ${\mathrm{NO}}_2$ are needed to form $75.0$ g of ${\mathrm{HNO}}_3$

Part (a) Step 1: Given information

We need to find the balanced equation. 

Part (a) Step 2: Explanation

We know that

The balanced equation is that in which the total mass of reactants is equal to the total mass of products.

Therefore, The balanced equation is, 

$3{\mathrm{NO}}_2+{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{HNO}}_3+\mathrm{NO}$

Part (b) Step 1: Given information

We need to find the no. of moles of each product are produced from $0.250$ mole of ${\mathrm{H}}_2\mathrm{O}$

Part (b) Step 2: Explanation

From part (a) 

We know that

The a mole of ${\mathrm{H}}_2\mathrm{O}$ produces style="max-width: none; vertical-align: -4px;" $2$ mole of ${\mathrm{HNO}}_3$ and a mole of $\mathrm{NO}$

Therefore, style="max-width: none; vertical-align: -4px;" $0.250$ mole of ${\mathrm{H}}_2\mathrm{O}$ produces $\frac{2 \mathrm{mole}\times 0.250 \cancel{\mathrm{mole}}}{\cancel{\mathrm{mole}}}=0.5 \mathrm{mole}$ of ${\mathrm{HNO}}_3$ and $\frac{1 \mathrm{mole}\times 0.250 \cancel{\mathrm{mole}}}{\cancel{\mathrm{mole}}} =0.25 \mathrm{mole}$ of $\mathrm{NO}$

Part (c) Step 1: Given information

We need to find the mass of ${\mathrm{HNO}}_3$ produced from $60.0$ g of ${\mathrm{NO}}_2$

Part (c) Step 2: Explanation

From part (a) 

We know that

The $46$ g of ${\mathrm{NO}}_2$  produces $63$ g of ${\mathrm{HNO}}_3$

Therefore, style="max-width: none; vertical-align: -4px;" $60$ g of ${\mathrm{NO}}_2$ produces $\frac{63\cancel{\mathrm{g} }}{46 \cancel{\mathrm{g}}}\times 60 \mathrm{g}=82.34 \mathrm{g}$ of ${\mathrm{HNO}}_3$

Part (d) Step 1: Given information

We need to find the mass of ${\mathrm{NO}}_2$ needed to form $75$ g of ${\mathrm{HNO}}_3$

Part (d) Step 2: Explanation

From part (a)

We know that

The $63$ g of ${\mathrm{HNO}}_3$ is formed from $46$ g of ${\mathrm{NO}}_2$

Therefore, $75$ g of ${\mathrm{HNO}}_3$ is formed from $\frac{46 \cancel{\mathrm{g}}}{63 \cancel{\mathrm{g}}}\times 75 \mathrm{g}=54.78 \mathrm{g}$ of ${\mathrm{NO}}_2$