- First, calculate the moles of \(O_2\) using the given mass and molar mass: \(n(O_2) = \frac{1.45 \times 10^{-3} \, \text{g}}{32.00 \, \text{g/mol}} = 4.53125 \times 10^{-5} \, \text{mol}\).
- Apply the stoichiometric ratio from the balanced equation (\(1\) mol \(O_2 : 2\) mol \(H_2O\)) to find the moles of \(H_2O\): \(n(H_2O) = 2 \times n(O_2) = 9.0625 \times 10^{-5} \, \text{mol}\).
- Calculate the mass of the produced \(H_2O\) by multiplying its moles by its molar mass: \(9.0625 \times 10^{-5} \, \text{mol} \times 18.02 \, \text{g/mol} \approx 1.63 \times 10^{-3} \, \text{g}\).
- The final answer is \(\boxed{1.63 \times 10^{-3} \, \text{g}}\).
Explanation
- Problem Analysis
To determine the mass of water formed, we begin by analyzing the chemical equation: \[2H_2 + O_2 \rightarrow 2H_2O\]
This balanced equation tells us that 1 mole of oxygen gas (\(O_2\)) reacts to produce 2 moles of water (\(H_2O\)). We are given the mass of oxygen as \(1.45 \times 10^{-3} \, \text{g}\) and the molar masses of \(O_2\) (\(32.00 \, \text{g/mol}\)) and \(H_2O\) (\(18.02 \, \text{g/mol}\)).
- Calculating Moles of Reactants and Products
First, calculate the number of moles of \(O_2\) present:
\[n(O_2) = \frac{\text{mass}(O_2)}{\text{molar mass}(O_2)} = \frac{1.45 \times 10^{-3} \, \text{g}}{32.00 \, \text{g/mol}} \approx 4.53125 \times 10^{-5} \, \text{mol}\]
Next, using the stoichiometric relationship from the balanced equation (1 mole of \(O_2\) produces 2 moles of \(H_2O\)), calculate the moles of \(H_2O\) produced:
\[n(H_2O) = 2 \times n(O_2) = 2 \times 4.53125 \times 10^{-5} \, \text{mol} = 9.0625 \times 10^{-5} \, \text{mol}\]
- Final Mass Calculation
Finally, determine the mass of the produced water by multiplying the moles of \(H_2O\) by its molar mass:
\[\text{mass}(H_2O) = n(H_2O) \times \text{molar mass}(H_2O)\]
\[\text{mass}(H_2O) = 9.0625 \times 10^{-5} \, \text{mol} \times 18.02 \, \text{g/mol} \approx 1.6330625 \times 10^{-3} \, \text{g}\]
Rounding this to significant figures, we get approximately \(1.63 \times 10^{-3} \, \text{g}\), which corresponds to option (c).
Examples
Stoichiometry is fundamental in industrial chemistry for determining how much reactant is needed to produce a specific amount of product, such as calculating the amount of hydrogen and oxygen required to power a hydrogen fuel cell vehicle. By understanding the molar ratios in chemical reactions, engineers ensure efficient use of resources and minimize waste, which is critical for sustainable production processes in medicine, energy, and material science. This process demonstrates how chemical balance dictates the efficiency of real-world energy transformations.