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]A. 1.12 dm3B. 2.24dm3C. 4.48dm3D. 8.96dm3
Question
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]A. 1.12 dm3B. 2.24dm3C. 4.48dm3D. 8.96dm3
Solution
To solve this problem, we need to follow these steps:
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First, we need to calculate the number of moles of zinc (Zn) that are reacting. We do this by dividing the mass of Zn by its molar mass. The molar mass of Zn is given as 65 g/mol. So, the number of moles of Zn is 3.25 g / 65 g/mol = 0.05 mol.
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From the balanced chemical equation, we can see that one mole of Zn produces one mole of H2 gas. Therefore, the number of moles of H2 gas produced is also 0.05 mol.
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At standard temperature and pressure (s.t.p.), one mole of any gas occupies a volume of 22.4 dm^3. Therefore, the volume of H2 gas produced is 0.05 mol * 22.4 dm^3/mol = 1.12 dm^3.
So, the answer is A. 1.12 dm^3.
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