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

Question

UpStudy AI · Accepted Answer

To solve the problem, we will follow the steps outlined in the questions.

### a. Balance the equation.

The reaction involving nitrogen ($$ \mathrm{N}_{2} $$) and hydrogen ($$ \mathrm{H}_{2} $$) to produce phosphorous acid ($$ \mathrm{H}_{3} \mathrm{PO}_{3} $$) is not standard. However, if we assume a general reaction of nitrogen and hydrogen forming ammonia ($$ \mathrm{NH}_{3} $$), we can write the balanced equation as follows:

$$
\mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3}
$$

### b. How many mol of $$ \mathrm{H}_{2} $$ would react with 1 mol of $$ \mathrm{N}_{2} $$?

From the balanced equation, we see that 1 mol of $$ \mathrm{N}_{2} $$ reacts with 3 mol of $$ \mathrm{H}_{2} $$.

### c. How many mol of product would form from 1 mol of $$ \mathrm{N}_{2} $$?

From the balanced equation, 1 mol of $$ \mathrm{N}_{2} $$ produces 2 mol of $$ \mathrm{NH}_{3} $$.

### 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 $$ \mathrm{N}_{2} $$.

First, we need to calculate the number of moles of $$ \mathrm{N}_{2} $$ in 14.0 g. The molar mass of $$ \mathrm{N}_{2} $$ is approximately 28.02 g/mol.

$$
\text{moles of } \mathrm{N}_{2} = \frac{14.0 \, \text{g}}{28.02 \, \text{g/mol}}
$$

Next, we will calculate the moles of $$ \mathrm{H}_{2} $$ required using the stoichiometry from part (b).

### e. For conditions outlined in part (d), how many g of product would form?

Using the moles of $$ \mathrm{N}_{2} $$ calculated in part (d), we can find the moles of $$ \mathrm{NH}_{3} $$ produced using the stoichiometry from part (c). Then, we will convert the moles of $$ \mathrm{NH}_{3} $$ to grams using its molar mass (approximately 17.03 g/mol).

Now, let's perform the calculations for parts (d) and (e).

First, we will calculate the moles of $$ \mathrm{N}_{2} $$ and then the moles of $$ \mathrm{H}_{2} $$ required. After that, we will calculate the grams of product formed.

Let's start with part (d).

Calculating the moles of $$ \mathrm{N}_{2} $$:

$$
\text{moles of } \mathrm{N}_{2} = \frac{14.0}{28.02}
$$

Then, calculating the moles of $$ \mathrm{H}_{2} $$ required:

$$
\text{moles of } \mathrm{H}_{2} = 3 \times \text{moles of } \mathrm{N}_{2}
$$

Next, calculating the moles of $$ \mathrm{NH}_{3} $$ produced:

$$
\text{moles of } \mathrm{NH}_{3} = 2 \times \text{moles of } \mathrm{N}_{2}
$$

Finally, converting moles of $$ \mathrm{NH}_{3} $$ to grams:

$$
\text{grams of } \mathrm{NH}_{3} = \text{moles of } \mathrm{NH}_{3} \times 17.03
$$

Now, I will perform these calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$3\left(\frac{14}{28.02}\right)$$
- step1: Divide the numbers:
  $$3\times \frac{700}{1401}$$
- step2: Reduce the numbers:
  $$1\times \frac{700}{467}$$
- step3: Multiply:
  $$\frac{700}{467}$$
Calculate or simplify the expression $$ 2*(14.0/28.02) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$2\left(\frac{14}{28.02}\right)$$
- step1: Divide the numbers:
  $$2\times \frac{700}{1401}$$
- step2: Multiply:
  $$\frac{2\times 700}{1401}$$
- step3: Multiply:
  $$\frac{1400}{1401}$$
Calculate or simplify the expression $$ 2*(14.0/28.02)*17.03 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$2\left(\frac{14}{28.02}\right)\times 17.03$$
- step1: Divide the numbers:
  $$2\times \frac{700}{1401}\times 17.03$$
- step2: Multiply the terms:
  $$\frac{1400}{1401}\times 17.03$$
- step3: Convert the expressions:
  $$\frac{1400}{1401}\times \frac{1703}{100}$$
- step4: Multiply the fractions:
  $$\frac{1400\times 1703}{1401\times 100}$$
- step5: Multiply the terms:
  $$\frac{2384200}{1401\times 100}$$
- step6: Multiply the terms:
  $$\frac{2384200}{140100}$$
- step7: Reduce the fraction:
  $$\frac{23842}{1401}$$
Calculate or simplify the expression $$ 14.0/28.02 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{14}{28.02}$$
- step1: Convert the expressions:
  $$\frac{14}{\frac{1401}{50}}$$
- step2: Multiply by the reciprocal:
  $$14\times \frac{50}{1401}$$
- step3: Multiply:
  $$\frac{14\times 50}{1401}$$
- step4: Multiply:
  $$\frac{700}{1401}$$
Let's summarize the calculations:

### d. Calculate the number of moles of $$ \mathrm{N}_{2} $$:
$$
\text{moles of } \mathrm{N}_{2} = \frac{14.0 \, \text{g}}{28.02 \, \text{g/mol}} \approx \frac{700}{1401} \text{ mol} \approx 0.499 \text{ mol}
$$

### Calculate the moles of $$ \mathrm{H}_{2} $$ required:
$$
\text{moles of } \mathrm{H}_{2} = 3 \times \text{moles of } \mathrm{N}_{2} \approx 3 \times \frac{700}{1401} \approx \frac{2100}{1401} \text{ mol} \approx 1.498 \text{ mol}
$$

### e. Calculate the moles of $$ \mathrm{NH}_{3} $$ produced:
$$
\text{moles of } \mathrm{NH}_{3} = 2 \times \text{moles of } \mathrm{N}_{2} \approx 2 \times \frac{700}{1401} \approx \frac{1400}{1401} \text{ mol} \approx 0.999 \text{ mol}
$$

### Convert moles of $$ \mathrm{NH}_{3} $$ to grams:
$$
\text{grams of } \mathrm{NH}_{3} = \text{moles of } \mathrm{NH}_{3} \times 17.03 \approx \frac{1400}{1401} \times 17.03 \approx \frac{23842}{1401} \text{ g} \approx 17.0 \text{ g}
$$

### Summary of Results:
- **Moles of $$ \mathrm{H}_{2} $$ required:** $$ \approx 1.498 \text{ mol} $$
- **Moles of $$ \mathrm{NH}_{3} $$ produced:** $$ \approx 0.999 \text{ mol} $$
- **Grams of $$ \mathrm{NH}_{3} $$ produced:** $$ \approx 17.0 \text{ g} $$

If you have any further questions or need additional assistance, feel free to ask!
To solve the problem: a. The balanced equation is $$ \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} $$. b. 1 mol of $$ \mathrm{N}_{2} $$ reacts with 3 mol of $$ \mathrm{H}_{2} $$. c. 1 mol of $$ \mathrm{N}_{2} $$ produces 2 mol of $$ \mathrm{NH}_{3} $$. d. For 14.0 g of $$ \mathrm{N}_{2} $$, 1.498 mol of $$ \mathrm{H}_{2} $$ is required. e. Under the given conditions, 17.0 g of $$ \mathrm{NH}_{3} $$ would form.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Chemistry Questions