4.91 How many mol of $$ \mathrm{H}_{3} \mathrm{PO}_{3} $$ are produced? 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

Question

UpStudy AI · Accepted Answer

**a. Balancing the equation**

We start with the unbalanced reaction
$$
\mathrm{N}_{2} (g) + \mathrm{H}_{2} (g) \longrightarrow \mathrm{HN}_{3} (g)
$$
The compound $$\mathrm{HN}_{3}$$ contains 1 H atom and 3 N atoms per molecule. Let the coefficients be $$a$$, $$b$$, and $$c$$ for $$\mathrm{N}_{2}$$, $$\mathrm{H}_{2}$$, and $$\mathrm{HN}_{3}$$ respectively:
$$
a\, \mathrm{N}_{2} + b\, \mathrm{H}_{2} \longrightarrow c\, \mathrm{HN}_{3}
$$
Balance the nitrogen atoms:
$$
2a = 3c \quad \Longrightarrow \quad a = \frac{3c}{2}
$$
Balance the hydrogen atoms:
$$
2b = c \quad \Longrightarrow \quad c = 2b
$$
Substitute $$c = 2b$$ into the relation for $$a$$:
$$
a = \frac{3(2b)}{2} = 3b
$$
Choose the smallest integer value for $$b$$, which is 1. Then:
$$
a = 3,\quad c = 2
$$
Thus, the balanced equation is:
$$
3\, \mathrm{N}_{2} + \mathrm{H}_{2} \longrightarrow 2\, \mathrm{HN}_{3}
$$

---

**b. Moles of $$\mathrm{H}_{2}$$ reacting with 1 mol of $$\mathrm{N}_{2}$$**

The balanced equation shows that:
$$
3\, \mathrm{N}_{2} : 1\, \mathrm{H}_{2}
$$
For 3 moles of $$\mathrm{N}_{2}$$, 1 mole of $$\mathrm{H}_{2}$$ is required. For 1 mole of $$\mathrm{N}_{2}$$:
$$
\mathrm{moles\ of\ } \mathrm{H}_{2} = \frac{1}{3} \text{ mol}
$$

---

**c. Moles of product formed from 1 mol of $$\mathrm{N}_{2}$$**

From the balanced equation:
$$
3\, \mathrm{N}_{2} \longrightarrow 2\, \mathrm{HN}_{3}
$$
This indicates that 3 moles of $$\mathrm{N}_{2}$$ produce 2 moles of $$\mathrm{HN}_{3}$$. Therefore, 1 mole of $$\mathrm{N}_{2}$$ produces:
$$
\mathrm{moles\ of\ } \mathrm{HN}_{3} = \frac{2}{3} \text{ mol}
$$

---

**d. Moles of $$\mathrm{H}_{2}$$ required for 14.0 g of $$\mathrm{N}_{2}$$**

1. First, calculate the moles of $$\mathrm{N}_{2}$$. The molar mass of $$\mathrm{N}_{2}$$ is:
$$
M(\mathrm{N}_{2}) = 14.0\, \mathrm{g/mol} \times 2 = 28.0\, \mathrm{g/mol}
$$
Thus,
$$
\mathrm{moles\ of\ } \mathrm{N}_{2} = \frac{14.0\, \mathrm{g}}{28.0\, \mathrm{g/mol}} = 0.5\, \text{mol}
$$

2. From part (b), we require $$\frac{1}{3}$$ mol of $$\mathrm{H}_{2}$$ per mole of $$\mathrm{N}_{2}$$. Therefore, for 0.5 mol of $$\mathrm{N}_{2}$$:
$$
\mathrm{moles\ of\ } \mathrm{H}_{2} = 0.5 \times \frac{1}{3} = \frac{0.5}{3} \approx 0.167\, \text{mol}
$$

---

**e. Grams of product formed for the given conditions**

First, determine the moles of product ($$\mathrm{HN}_{3}$$) produced from 0.5 mol of $$\mathrm{N}_{2}$$.

The balanced equation shows:
$$
3\, \mathrm{N}_{2} \longrightarrow 2\, \mathrm{HN}_{3}
$$
Thus, the mole ratio is:
$$
\mathrm{\frac{HN_3}{N_2}} = \frac{2}{3}
$$
So,
$$
\mathrm{moles\ of\ } \mathrm{HN}_{3} = 0.5 \times \frac{2}{3} \approx 0.333\, \text{mol}
$$

Now, calculate the molar mass of $$\mathrm{HN}_{3}$$:
$$
M(\mathrm{HN}_{3}) = M(\mathrm{H}) + 3 \times M(\mathrm{N}) = 1.0\, \mathrm{g/mol} + 3(14.0\, \mathrm{g/mol}) = 1.0 + 42.0 = 43.0\, \mathrm{g/mol}
$$

Finally, find the mass of product:
$$
\text{Mass of } \mathrm{HN}_{3} = 0.333\, \text{mol} \times 43.0\, \mathrm{g/mol} \approx 14.33\, \mathrm{g}
$$
a. The balanced equation is: $$ 3\, \mathrm{N}_{2} + \mathrm{H}_{2} \longrightarrow 2\, \mathrm{HN}_{3} $$ b. 0.333 mol of $$\mathrm{H}_{2}$$ react with 1 mol of $$\mathrm{N}_{2}$$. c. 0.667 mol of $$\mathrm{HN}_{3}$$ are produced from 1 mol of $$\mathrm{N}_{2}$$. d. 0.167 mol of $$\mathrm{H}_{2}$$ are required to react with 14.0 g of $$\mathrm{N}_{2}$$. e. Approximately 14.33 g of $$\mathrm{HN}_{3}$$ are formed.

Solución de inteligencia artificial de Upstudy

Responder

Solución

preguntas relacionadas

Latest Chemistry Questions