What volume of hydrogen gas is produced when 0.829 mol of iron reacts completely

Question

What volume of hydrogen gas is produced when 0.829 mol of iron reacts completely according to the following reaction at 0oC and 1 atm?

iron (s) + hydrochloric acid (aq) iron(II) chloride (aq) + hydrogen (g)


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Explanation / Answer

What volume of hydrogen gas is produced when 0.829 mol of iron reacts completely according to the following reaction at 0oC and 1 atm? iron (s) + hydrochloric acid (aq) iron(II) chloride (aq) + hydrogen (g) ANSWER: Fe(s) +2HCl(aq) --------------> FeCl2(aq) + H2(g) Step 1: Use the moles ratio of Fe:H2, find the moles of H2. The moles of H2 = 0.829 mol x(1mol H2/1mole Fe) = 0.829 moles of H2. Step 2: Ideal Gas Law, PV = nRT Where P is Pressure, 1 atm, T is temperature, 0oC or 273 K n is number of moles, 0.829, R is Gas constant, 0.0821 atm L/mol K Volume (V) = nRT/P = [(0.829 moles)x(0.08206 atm L/mol K)x(273 K)]/(1atm) = 18.57 L So the volume of H2 gas is 18.57 L Note: We can also use the data of STP condition because the Pressure is 1 atm and Temperature is 0oC. At STP condition, 1 moles gas = 22.4 L So 0.829 moles H2 gas = 0.829 mole x(22.4 L/mol) = 18.57 L

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