Question
8.08 g of hydrogen gas reacting with excess oxygen gas will produce how many grams of water?
Answers
GPT-4o mini
To find out how many grams of water are produced from the reaction of hydrogen gas with excess oxygen gas, we first need to write the balanced chemical equation for the reaction:
\[
2 \text{H}_2 (g) + \text{O}_2 (g) \rightarrow 2 \text{H}_2\text{O} (l)
\]
From the balanced equation, we can see that 2 moles of hydrogen gas (\( \text{H}_2 \)) produce 2 moles of water (\( \text{H}_2\text{O} \)). This means that the mole ratio of hydrogen gas to water is 1:1.
Next, we need to convert grams of hydrogen gas to moles. The molar mass of hydrogen gas (\( \text{H}_2 \)) is approximately 2.02 g/mol.
Now, we calculate the number of moles of hydrogen gas in 8.08 grams:
\[
\text{Number of moles of } \text{H}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{8.08 \text{ g}}{2.02 \text{ g/mol}} \approx 4.00 \text{ mol}
\]
Since the mole ratio of hydrogen to water is 1:1, 4.00 moles of hydrogen will produce 4.00 moles of water.
Next, we calculate the mass of water produced. The molar mass of water (\( \text{H}_2\text{O} \)) is approximately 18.02 g/mol.
Now we calculate the mass of water produced:
\[
\text{Mass of water} = \text{number of moles} \times \text{molar mass} = 4.00 \text{ mol} \times 18.02 \text{ g/mol} \approx 72.08 \text{ g}
\]
Therefore, 8.08 grams of hydrogen gas reacting with excess oxygen gas will produce approximately **72.08 grams of water**.
\[
2 \text{H}_2 (g) + \text{O}_2 (g) \rightarrow 2 \text{H}_2\text{O} (l)
\]
From the balanced equation, we can see that 2 moles of hydrogen gas (\( \text{H}_2 \)) produce 2 moles of water (\( \text{H}_2\text{O} \)). This means that the mole ratio of hydrogen gas to water is 1:1.
Next, we need to convert grams of hydrogen gas to moles. The molar mass of hydrogen gas (\( \text{H}_2 \)) is approximately 2.02 g/mol.
Now, we calculate the number of moles of hydrogen gas in 8.08 grams:
\[
\text{Number of moles of } \text{H}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{8.08 \text{ g}}{2.02 \text{ g/mol}} \approx 4.00 \text{ mol}
\]
Since the mole ratio of hydrogen to water is 1:1, 4.00 moles of hydrogen will produce 4.00 moles of water.
Next, we calculate the mass of water produced. The molar mass of water (\( \text{H}_2\text{O} \)) is approximately 18.02 g/mol.
Now we calculate the mass of water produced:
\[
\text{Mass of water} = \text{number of moles} \times \text{molar mass} = 4.00 \text{ mol} \times 18.02 \text{ g/mol} \approx 72.08 \text{ g}
\]
Therefore, 8.08 grams of hydrogen gas reacting with excess oxygen gas will produce approximately **72.08 grams of water**.