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Multiple Choice
What volume of oxygen gas, measured at STP, is required to completely react with 4.00 g of hydrogen gas (H_2) according to the reaction: 2 H_2(g) + O_2(g) → 2 H_2O(g)?
A
5.60 L
B
44.8 L
C
22.4 L
D
11.2 L
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Verified step by step guidance
1
Write down the balanced chemical equation: \(2\ H_2(g) + O_2(g) \rightarrow 2\ H_2O(g)\).
Calculate the number of moles of hydrogen gas (\(H_2\)) using its molar mass. Use the formula: \(\text{moles of } H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2}\). The molar mass of \(H_2\) is approximately 2.02 g/mol.
Use the mole ratio from the balanced equation to find the moles of oxygen gas (\(O_2\)) required. According to the equation, 2 moles of \(H_2\) react with 1 mole of \(O_2\), so \(\text{moles of } O_2 = \frac{1}{2} \times \text{moles of } H_2\).
Recall that at STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters. Use this to calculate the volume of \(O_2\) gas: \(\text{volume of } O_2 = \text{moles of } O_2 \times 22.4\ L/mol\).
Combine all the steps to find the volume of oxygen gas required to completely react with 4.00 g of hydrogen gas.