1.00 g of magnesium ribbon reacted with 25.0 cm3 of 1.00 mol dm–3 hydrochloric acid. Determine the volume, in cm3, of a balloon filled with the hydrogen gas produced at SATP (25°C and 100 kPa). Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)1018 cm3619 cm3450 cm3310 cm3
Question
1.00 g of magnesium ribbon reacted with 25.0 cm3 of 1.00 mol dm–3 hydrochloric acid. Determine the volume, in cm3, of a balloon filled with the hydrogen gas produced at SATP (25°C and 100 kPa). Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)1018 cm3619 cm3450 cm3310 cm3
Solution
To solve this problem, we need to follow these steps:
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First, we need to determine the number of moles of magnesium. We know that the molar mass of magnesium (Mg) is approximately 24.31 g/mol. So, 1.00 g of Mg is approximately 1.00 g / 24.31 g/mol = 0.0411 mol.
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From the balanced chemical equation, we know that 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H2. Therefore, 0.0411 mol of Mg will produce 0.0411 mol of H2.
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At SATP (Standard Ambient Temperature and Pressure), which is 25°C and 100 kPa, 1 mole of any gas occupies 24.8 L or 24800 cm3. Therefore, 0.0411 mol of H2 will occupy 0.0411 mol * 24800 cm3/mol = 1018 cm3.
So, the volume of the balloon filled with the hydrogen gas produced at SATP is 1018 cm3.
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