To solve this problem, we need to follow these steps:

1. **Write the balanced chemical equation for the reform reaction:**
$$
\text{CH}_4(g) + \text{H}_2\text{O}(g) \rightarrow \text{CO}(g) + 3\text{H}_2(g)
$$

2. **Use the Ideal Gas Law to find the number of moles of methane consumed per second:**
The Ideal Gas Law is given by:
$$
PV = nRT
$$
where:
- $ P $ is the pressure (0.70 atm)
- $ V $ is the volume (986 L/s)
- $ R $ is the ideal gas constant (0.0821 L·atm/(K·mol))
- $ T $ is the temperature in Kelvin (191°C + 273.15 = 464.15 K)

Rearrange the Ideal Gas Law to solve for $ n $:
$$
n = \frac{PV}{RT}
$$

Plug in the values:
$$
n = \frac{(0.70 \text{ atm})(986 \text{ L/s})}{(0.0821 \text{ L·atm/(K·mol)})(464.15 \text{ K})}
$$

Calculate $ n $:
$$
n \approx \frac{690.2}{38.1} \approx 18.1 \text{ mol/s}
$$

3. **Determine the rate of dihydrogen production:**
From the balanced equation, 1 mole of CH4 produces 3 moles of H2. Therefore, the rate of H2 production is:
$$
3 \times 18.1 \text{ mol/s} = 54.3 \text{ mol/s}
$$

4. **Convert the rate of dihydrogen production to kilograms per second:**
The molar mass of H2 is approximately 2.02 g/mol. Convert moles per second to grams per second:
$$
54.3 \text{ mol/s} \times 2.02 \text{ g/mol} = 109.7 \text{ g/s}
$$

Convert grams per second to kilograms per second:
$$
109.7 \text{ g/s} = 0.1097 \text{ kg/s}
$$

5. **Round the answer to 2 significant digits:**
$$
0.11 \text{ kg/s}
$$

Therefore, the rate at which dihydrogen is being produced is $ 0.11 $ kilograms per second.

Question

To solve this problem, we need to follow these steps:

1. **Write the balanced chemical equation for the reform reaction:**
   $$
   \text{CH}_4(g) + \text{H}_2\text{O}(g) \rightarrow \text{CO}(g) + 3\text{H}_2(g)
   $$

2. **Use the Ideal Gas Law to find the number of moles of methane consumed per second:**
   The Ideal Gas Law is given by:
   $$
   PV = nRT
   $$
   where:
   - $ P $ is the pressure (0.70 atm)
   - $ V $ is the volume (986 L/s)
   - $ R $ is the ideal gas constant (0.0821 L·atm/(K·mol))
   - $ T $ is the temperature in Kelvin (191°C + 273.15 = 464.15 K)

Rearrange the Ideal Gas Law to solve for $ n $:
   $$
   n = \frac{PV}{RT}
   $$

Plug in the values:
   $$
   n = \frac{(0.70 \text{ atm})(986 \text{ L/s})}{(0.0821 \text{ L·atm/(K·mol)})(464.15 \text{ K})}
   $$

Calculate $ n $:
   $$
   n \approx \frac{690.2}{38.1} \approx 18.1 \text{ mol/s}
   $$

3. **Determine the rate of dihydrogen production:**
   From the balanced equation, 1 mole of CH4 produces 3 moles of H2. Therefore, the rate of H2 production is:
   $$
   3 \times 18.1 \text{ mol/s} = 54.3 \text{ mol/s}
   $$

4. **Convert the rate of dihydrogen production to kilograms per second:**
   The molar mass of H2 is approximately 2.02 g/mol. Convert moles per second to grams per second:
   $$
   54.3 \text{ mol/s} \times 2.02 \text{ g/mol} = 109.7 \text{ g/s}
   $$

Convert grams per second to kilograms per second:
   $$
   109.7 \text{ g/s} = 0.1097 \text{ kg/s}
   $$

5. **Round the answer to 2 significant digits:**
   $$
   0.11 \text{ kg/s}
   $$

Therefore, the rate at which dihydrogen is being produced is $ 0.11 $ kilograms per second.

Knowee AI · Accepted Answer