Consider the reaction represented by the following equation: Zn(s) + 2HCI(aq) → ZnCI2(aq) + H2(g) what volume of hydrogen gas is produced at s.t.p. when 3.2...
Question 1 Report
Consider the reaction represented by the following equation: Zn(s) + 2HCI(aq) → ZnCI2(aq) + H2(g) what volume of hydrogen gas is produced at s.t.p. when 3.25 g of zinc reacts with excess dilute HCI? [Zn = 65, Molar gas volume at s.t.p. = 22.4 dm-3]
Answer Details
First, we need to calculate the number of moles of zinc used in the reaction. Moles of Zn = mass ÷ molar mass Moles of Zn = 3.25 g ÷ 65 g/mol Moles of Zn = 0.05 mol From the balanced equation, we know that 1 mole of Zn produces 1 mole of H2 gas. Therefore, the number of moles of H2 gas produced is also 0.05 mol. At s.t.p (standard temperature and pressure), one mole of any gas occupies 22.4 dm3. So, the volume of hydrogen gas produced at s.t.p is: Volume = number of moles × molar gas volume at s.t.p Volume = 0.05 mol × 22.4 dm-3/mol Volume = 1.12 dm3 Therefore, the answer is 1.12 dm3.
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