2H2 +O2 → 2H2O
What mass of water forms when 1.45 × 10-3 g O2 react completely? (Molar mass of O2 = 32.00 g/mol; molar mass of H2O = 18.02 g/mol)
1.63 × 10-3 g
8.16 × 10-4 g
1.29 × 10-3 g
The Correct Answer and Explanation is :
To determine the mass of water ((H_2O)) that forms when 1.45 × 10^-3 g of (O_2) reacts completely, we start by using the balanced chemical equation:
[
2H_2 + O_2 \rightarrow 2H_2O
]
From this equation, we see that one mole of (O_2) produces two moles of (H_2O).
Step 1: Convert the mass of (O_2) to moles.
To convert grams of (O_2) to moles, we use the formula:
[
\text{moles} = \frac{\text{mass}}{\text{molar mass}}
]
The molar mass of (O_2) is 32.00 g/mol. Thus,
[
\text{moles of } O_2 = \frac{1.45 \times 10^{-3} \text{ g}}{32.00 \text{ g/mol}} \approx 4.53 \times 10^{-5} \text{ mol}
]
Step 2: Use the stoichiometry of the reaction to find moles of (H_2O).
According to the balanced equation, 1 mole of (O_2) produces 2 moles of (H_2O). Therefore, the moles of (H_2O) produced from 4.53 × 10^-5 moles of (O_2) is:
[
\text{moles of } H_2O = 2 \times (4.53 \times 10^{-5}) \approx 9.06 \times 10^{-5} \text{ mol}
]
Step 3: Convert moles of (H_2O) to grams.
Now, we convert moles of (H_2O) to grams using its molar mass (18.02 g/mol):
[
\text{mass of } H_2O = \text{moles} \times \text{molar mass} = 9.06 \times 10^{-5} \text{ mol} \times 18.02 \text{ g/mol} \approx 1.63 \times 10^{-3} \text{ g}
]
Conclusion:
The mass of water that forms when 1.45 × 10^-3 g of (O_2) reacts completely is 1.63 × 10^-3 g. This value is obtained through stoichiometric calculations, ensuring that we accurately account for the mole ratios involved in the reaction.