Question
8.08 g of hydrogen gas reacting with 32.0 g of oxygen gas will produce how many grams of water?
Answers
GPT-4o mini
To calculate the amount of water produced from the reaction between hydrogen gas and oxygen gas, we first need to use the balanced chemical equation for the reaction:
\[
2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}
\]
Next, we need to determine the number of moles of hydrogen and oxygen involved in the reaction:
1. **Calculate moles of hydrogen (H₂)**:
- Molar mass of H₂ = 2.02 g/mol (1.01 g/mol for each hydrogen atom, and there are 2 atoms)
- Moles of H₂ = mass (g) / molar mass (g/mol)
\[
\text{Moles of H}_2 = \frac{8.08 \, \text{g}}{2.02 \, \text{g/mol}} \approx 4.0 \, \text{moles}
\]
2. **Calculate moles of oxygen (O₂)**:
- Molar mass of O₂ = 32.00 g/mol
- Moles of O₂ = mass (g) / molar mass (g/mol)
\[
\text{Moles of O}_2 = \frac{32.0 \, \text{g}}{32.00 \, \text{g/mol}} = 1.0 \, \text{mole}
\]
3. **Determine the limiting reactant**:
From the balanced equation, 2 moles of H₂ react with 1 mole of O₂. Therefore, we need 2 moles of H₂ for every mole of O₂.
- For 1 mole of O₂, we need:
\[
2 \, \text{moles of H}_2
\]
From our calculations, we have 4.0 moles of H₂ and 1.0 mole of O₂. Oxygen is the limiting reactant because we have more than enough H₂ to completely react with the available O₂.
4. **Calculate moles of water (H₂O) produced**:
According to the equation, 1 mole of O₂ produces 2 moles of H₂O. Therefore, from 1 mole of O₂, we can produce:
\[
\text{Moles of H}_2\text{O} = 2 \times \text{Moles of O}_2 = 2 \times 1.0 \, \text{mole} = 2.0 \, \text{moles}
\]
5. **Calculate the mass of water produced**:
- Molar mass of H₂O = 18.02 g/mol
\[
\text{Mass of H}_2\text{O} = \text{moles} \times \text{molar mass}
\]
\[
\text{Mass of H}_2\text{O} = 2.0 \, \text{moles} \times 18.02 \, \text{g/mol} = 36.04 \, \text{g}
\]
Therefore, the reaction of 8.08 g of hydrogen gas with 32.0 g of oxygen gas will produce approximately **36.04 grams of water**.
\[
2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}
\]
Next, we need to determine the number of moles of hydrogen and oxygen involved in the reaction:
1. **Calculate moles of hydrogen (H₂)**:
- Molar mass of H₂ = 2.02 g/mol (1.01 g/mol for each hydrogen atom, and there are 2 atoms)
- Moles of H₂ = mass (g) / molar mass (g/mol)
\[
\text{Moles of H}_2 = \frac{8.08 \, \text{g}}{2.02 \, \text{g/mol}} \approx 4.0 \, \text{moles}
\]
2. **Calculate moles of oxygen (O₂)**:
- Molar mass of O₂ = 32.00 g/mol
- Moles of O₂ = mass (g) / molar mass (g/mol)
\[
\text{Moles of O}_2 = \frac{32.0 \, \text{g}}{32.00 \, \text{g/mol}} = 1.0 \, \text{mole}
\]
3. **Determine the limiting reactant**:
From the balanced equation, 2 moles of H₂ react with 1 mole of O₂. Therefore, we need 2 moles of H₂ for every mole of O₂.
- For 1 mole of O₂, we need:
\[
2 \, \text{moles of H}_2
\]
From our calculations, we have 4.0 moles of H₂ and 1.0 mole of O₂. Oxygen is the limiting reactant because we have more than enough H₂ to completely react with the available O₂.
4. **Calculate moles of water (H₂O) produced**:
According to the equation, 1 mole of O₂ produces 2 moles of H₂O. Therefore, from 1 mole of O₂, we can produce:
\[
\text{Moles of H}_2\text{O} = 2 \times \text{Moles of O}_2 = 2 \times 1.0 \, \text{mole} = 2.0 \, \text{moles}
\]
5. **Calculate the mass of water produced**:
- Molar mass of H₂O = 18.02 g/mol
\[
\text{Mass of H}_2\text{O} = \text{moles} \times \text{molar mass}
\]
\[
\text{Mass of H}_2\text{O} = 2.0 \, \text{moles} \times 18.02 \, \text{g/mol} = 36.04 \, \text{g}
\]
Therefore, the reaction of 8.08 g of hydrogen gas with 32.0 g of oxygen gas will produce approximately **36.04 grams of water**.