Calculate the total volume of hydrogen gas that was liberated at S.T.P conditions (Zn = 65.4, molar gas volume = 22.4 litres at S.T.P) assuming one mole of zinc reacted.

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Asked on April 24, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-04-24T22:42:07.222Z

Here are the solutions to your questions. b) Calculate the total volume of hydrogen gas that was liberated at S.T.P conditions (Zn = 65.4, molar gas volume = 22.4 litres at S.T.P) To calculate the volume of hydrogen gas liberated, we need to know the amount of zinc that reacted. Since no specific mass of zinc is provided, we will assume that one mole of zinc reacts. Step 1: Write the balanced chemical equation for the reaction of zinc with an acid to produce hydrogen gas. A common reaction is zinc reacting with hydrochloric acid: Zn(s) + 2HCl(aq) → ZnCl_2(aq) + H_2(g) From the stoichiometry, 1 mole of Zn produces 1 mole of H_2. Step 2: Determine the moles of H_2 liberated. Assuming 1 mole of Zn reacts, then 1 mole of H_2 is produced. Moles of H_2 = 1 mol Step 3: Calculate the volume of H_2 at S.T.P. At S.T.P., 1 mole of any gas occupies 22.4 litres. Volume of H_2 = Moles of H_2 × Molar gas volume at S.T.P. Volume of H_2 = 1 mol × 22.4 litres/mol Volume of H_2 = 22.4 litres The total volume of hydrogen gas liberated, assuming one mole of zinc reacted, is 22.4 litres. 2. Calculate the mass of nitrogen (IV) oxide gas that would occupy the same volume as 10g of hydrogen gas at the same temperature and pressure. (H = 1.0, N = 14.0, O = 16.0) Step 1: Calculate the molar mass of hydrogen gas (H_2). Molar mass of H_2 = 2 × Atomic mass of H Molar mass of H_2 = 2 × 1.0 g/mol = 2.0 g/mol Step 2: Calculate the moles of hydrogen gas (H_2). Moles of H_2 = Mass of H_2Molar mass of H_2 Moles of H_2 = 10 g2.0 g/mol = 5.0 mol Step 3: Determine the moles of nitrogen (IV) oxide gas (NO_2). According to Avogadro's Law, at the same temperature and pressure, equal volumes of all gases contain the same number of moles. Since nitrogen (IV) oxide gas occupies the same volume as hydrogen gas at the same temperature and pressure, they must have the same number of moles. Moles of NO_2 = Moles of H_2 = 5.0 mol Step 4: Calculate the molar mass of nitrogen (IV) oxide gas (NO_2). Molar mass of NO_2 = Atomic mass of N + (2 × Atomic mass of O) Molar mass of NO_2 = 14.0 g/mol + (2 × 16.0 g/mol) Molar mass of NO_2 = 14.0 g/mol + 32.0 g/mol = 46.0 g/mol Step 5: Calculate the mass of nitrogen (IV) oxide gas (NO_2). Mass of NO_2 = Moles of NO_2 × Molar mass of NO_2 Mass of NO_2 = 5.0 mol × 46.0 g/mol Mass of NO_2 = 230 g The mass of nitrogen (IV) oxide gas is 230 g. Send me the next one 📸

Accepted Answer · 2026-04-24T22:42:07.222Z

Here are the solutions to your questions. b) Calculate the total volume of hydrogen gas that was liberated at S.T.P conditions (Zn = 65.4, molar gas volume = 22.4 litres at S.T.P) To calculate the volume of hydrogen gas liberated, we need to know the amount of zinc that reacted. Since no specific mass of zinc is provided, we will assume that one mole of zinc reacts. Step 1: Write the balanced chemical equation for the reaction of zinc with an acid to produce hydrogen gas. A common reaction is zinc reacting with hydrochloric acid: Zn(s) + 2HCl(aq) → ZnCl_2(aq) + H_2(g) From the stoichiometry, 1 mole of Zn produces 1 mole of H_2. Step 2: Determine the moles of H_2 liberated. Assuming 1 mole of Zn reacts, then 1 mole of H_2 is produced. Moles of H_2 = 1 mol Step 3: Calculate the volume of H_2 at S.T.P. At S.T.P., 1 mole of any gas occupies 22.4 litres. Volume of H_2 = Moles of H_2 × Molar gas volume at S.T.P. Volume of H_2 = 1 mol × 22.4 litres/mol Volume of H_2 = 22.4 litres The total volume of hydrogen gas liberated, assuming one mole of zinc reacted, is 22.4 litres. 2. Calculate the mass of nitrogen (IV) oxide gas that would occupy the same volume as 10g of hydrogen gas at the same temperature and pressure. (H = 1.0, N = 14.0, O = 16.0) Step 1: Calculate the molar mass of hydrogen gas (H_2). Molar mass of H_2 = 2 × Atomic mass of H Molar mass of H_2 = 2 × 1.0 g/mol = 2.0 g/mol Step 2: Calculate the moles of hydrogen gas (H_2). Moles of H_2 = Mass of H_2Molar mass of H_2 Moles of H_2 = 10 g2.0 g/mol = 5.0 mol Step 3: Determine the moles of nitrogen (IV) oxide gas (NO_2). According to Avogadro's Law, at the same temperature and pressure, equal volumes of all gases contain the same number of moles. Since nitrogen (IV) oxide gas occupies the same volume as hydrogen gas at the same temperature and pressure, they must have the same number of moles. Moles of NO_2 = Moles of H_2 = 5.0 mol Step 4: Calculate the molar mass of nitrogen (IV) oxide gas (NO_2). Molar mass of NO_2 = Atomic mass of N + (2 × Atomic mass of O) Molar mass of NO_2 = 14.0 g/mol + (2 × 16.0 g/mol) Molar mass of NO_2 = 14.0 g/mol + 32.0 g/mol = 46.0 g/mol Step 5: Calculate the mass of nitrogen (IV) oxide gas (NO_2). Mass of NO_2 = Moles of NO_2 × Molar mass of NO_2 Mass of NO_2 = 5.0 mol × 46.0 g/mol Mass of NO_2 = 230 g The mass of nitrogen (IV) oxide gas is 230 g. Send me the next one 📸

Calculate the total volume of hydrogen gas that was liberated at S.T.P conditions (Zn = 65.4, molar gas volume = 22.4 litres at S.T.P) assuming one mole of zinc reacted.

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Calculate the total volume of hydrogen gas that was liberated at S.T.P conditions (Zn = 65.4, molar gas volume = 22.4 litres at S.T.P) assuming one mole of zinc reacted.

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