4.91 For the reaction $$ \mathrm{N}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HN}_{3}(\mathrm{~g}) $$ a. Balance the equation. b. How many mol of $$ \mathrm{H}_{2} $$ would react with 1 mol of $$ \mathrm{N}_{2} $$ ? c. How many mol of product would form from 1 mol of $$ \mathrm{N}_{2} $$ ? d. If 14.0 g of $$ \mathrm{N}_{2} $$ were initially present, calculate the number of mol of $$ \mathrm{H}_{2} $$ required to react with all of the $$ \mathrm{N}_{2} $$. e. For conditions outlined in part (d), how many g of product would form?

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**(a)**  
We balance the equation by letting the coefficients be $$ a $$ for $$\mathrm{N}_2$$, $$ b $$ for $$\mathrm{H}_2$$, and $$ c $$ for $$\mathrm{HN}_3$$.

- Nitrogen balance:  
  $$
  2a = 3c
  $$
- Hydrogen balance:  
  $$
  2b = 1 \times c
  $$

Choose the smallest integers that satisfy these relationships. Let  
$$
a=3
$$
Then, from nitrogen balance:  
$$
2(3)=3c \implies 6=3c \implies c=2
$$
And from hydrogen balance:  
$$
2b = c \implies 2b=2 \implies b=1
$$

Thus, the balanced equation is:  
$$
3\,\mathrm{N}_2(\mathrm{g})+\mathrm{H}_2(\mathrm{g})\rightarrow 2\,\mathrm{HN}_3(\mathrm{g})
$$

**(b)**  
From the balanced equation, $$3$$ moles of $$\mathrm{N}_2$$ require $$1$$ mole of $$\mathrm{H}_2$$. Therefore for $$1$$ mole of $$\mathrm{N}_2$$:
$$
\mathrm{mol\ H_2} = 1 \times \left(\frac{1}{3}\right)=\frac{1}{3}\text{ mol}
$$

**(c)**  
From the balanced equation, $$3$$ moles of $$\mathrm{N}_2$$ produce $$2$$ moles of $$\mathrm{HN}_3$$. Thus, for $$1$$ mole of $$\mathrm{N}_2$$:
$$
\mathrm{mol\ HN_3} = 1 \times \left(\frac{2}{3}\right)=\frac{2}{3}\text{ mol}
$$

**(d)**  
First, calculate the number of moles in 14.0 g of $$\mathrm{N}_2$$. The molar mass of $$\mathrm{N}_2$$ is  
$$
M(\mathrm{N}_2)= 2(14.0\,\mathrm{g/mol})=28.0\,\mathrm{g/mol}
$$

So, the moles of $$\mathrm{N}_2$$ are:  
$$
\mathrm{mol\ N_2}=\frac{14.0\,\mathrm{g}}{28.0\,\mathrm{g/mol}}=0.5\,\text{mol}
$$

According to the balanced equation, $$3\,\mathrm{N}_2$$ requires $$1\,\mathrm{H}_2$$. Therefore, for 0.5 mol $$\mathrm{N}_2$$:
$$
\mathrm{mol\ H_2} = 0.5 \times \left(\frac{1}{3}\right)=\frac{0.5}{3}\approx0.167\,\text{mol}
$$

**(e)**  
From the balanced equation, $$3\,\mathrm{N}_2$$ produces $$2\,\mathrm{HN}_3$$, so for 0.5 mol $$\mathrm{N}_2$$:
$$
\mathrm{mol\ HN_3} = 0.5 \times \left(\frac{2}{3}\right)=\frac{1}{3}\,\text{mol}\approx0.333\,\text{mol}
$$

Next, calculate the mass of $$\mathrm{HN}_3$$ produced. The molar mass of $$\mathrm{HN}_3$$ is:  
$$
M(\mathrm{HN}_3)= 1.0\,\mathrm{g/mol} + 3(14.0\,\mathrm{g/mol})=1.0+42.0=43.0\,\mathrm{g/mol}
$$

Thus, the mass is:  
$$
\text{mass}=\left(\frac{1}{3}\,\text{mol}\right)(43.0\,\mathrm{g/mol})\approx14.3\,\mathrm{g}
$$
**(a)** Balanced equation: $$ 3\,\mathrm{N}_2(\mathrm{g}) + \mathrm{H}_2(\mathrm{g}) \rightarrow 2\,\mathrm{HN}_3(\mathrm{g}) $$ **(b)** 1 mole of $$\mathrm{N}_2$$ reacts with $$\frac{1}{3}$$ mole of $$\mathrm{H}_2$$. **(c)** 1 mole of $$\mathrm{N}_2$$ produces $$\frac{2}{3}$$ mole of $$\mathrm{HN}_3$$. **(d)** 0.5 mole of $$\mathrm{N}_2$$ requires approximately 0.167 mole of $$\mathrm{H}_2$$. **(e)** 0.5 mole of $$\mathrm{N}_2$$ produces approximately 14.3 grams of $$\mathrm{HN}_3$$.

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