When 4.0 g H2 reacts with excess O2 (2H2+O2-2H2O), grams of water? show steps

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When 4.0 g H2 reacts with excess O2 (2H2+O2->2H2O), grams of water? show steps

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When 4.0 g of H2 reacts with excess O2, 36 g of water are produced. Steps: 1. Calculate moles of H2: Molar mass of H2 = 2 × 1.008 g/mol = 2.016 g/mol; Moles of H2 = Mass / Molar mass = 4.0 g / 2.016 g/mol = 1.984 mol H2 2. Use stoichiometry to find moles of H2O: From the balanced equation (2H2 + O2 -> 2H2O), 2 moles of H2 produce 2 moles of H2O. Moles of H2O = Moles of H2 * (2 mol H2O / 2 mol H2) = 1.984 mol H2O 3. Calculate mass of H2O: Molar mass of H2O = (2 × 1.008 g/mol) + 15.999 g/mol = 18.015 g/mol; Mass of H2O = Moles of H2O * Molar mass = 1.984 mol * 18.015 g/mol = 35.74 g 4. Round to appropriate significant figures: The initial mass (4.0 g) has two significant figures, so the final answer should also have two significant figures. Mass of H2O = 36 g

When 4.0 g H2 reacts with excess O2 (2H2+O2->2H2O), grams of water? show steps

Answer

Step-by-step solution

Calculate moles of H2

Use stoichiometry to find moles of H2O

Calculate mass of H2O

Round to appropriate significant figures

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