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

Question 1

Answer · 2026-06-04T17:37:24.782Z

When 4.0 g of H2 reacts with excess O2, 36 g of water are produced. Steps: 1. Calculate the 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 the mole ratio from the balanced equation to find moles of H2O: The balanced equation is 2H2 + O2 -> 2H2O. From the equation, 2 moles of H2 produce 2 moles of H2O. Moles of H2O = 1.984 mol H2 * (2 mol H2O / 2 mol H2) = 1.984 mol H2O 3. Calculate the mass of H2O: Molar mass of H2O = (2 1.008 g/mol) + (1 15.999 g/mol) = 18.015 g/mol Mass of H2O = Moles Molar mass = 1.984 mol 18.015 g/mol = 35.74 g H2O Rounding to two significant figures (due to 4.0 g H2), the mass of water produced is 36 g.

Question 2

When 4.0 g of H2 reacts with excess O2, 36 g of water are produced. Steps: 1. Calculate the 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 the mole ratio from the balanced equation to find moles of H2O: The balanced equation is 2H2 + O2 -> 2H2O. From the equation, 2 moles of H2 produce 2 moles of H2O. Moles of H2O = 1.984 mol H2 * (2 mol H2O / 2 mol H2) = 1.984 mol H2O 3. Calculate the mass of H2O: Molar mass of H2O = (2 1.008 g/mol) + (1 15.999 g/mol) = 18.015 g/mol Mass of H2O = Moles Molar mass = 1.984 mol 18.015 g/mol = 35.74 g H2O Rounding to two significant figures (due to 4.0 g H2), the mass of water produced is 36 g.

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

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