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3 - (chemiii. hi. summer,2017 (3) - stroichiometric relationship what is the@pression)for the volume of hydrogen gas, in dm^3 produced

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3 - (CHEMIII. HI. Summer,2017 (3) - Stroichiometric Relationship What is the@pression)for the volume of hydrogen gas, in dm^3 produced at STP when 0.30 g of magnesium reacts with excess hydrochloric acid solution? Mg(s)+2HCl(aq)arrow MgCl_(2)(aq)+H_(2)(g) Molar volume of an ideal gas at STP=22.7dm^3mol^-1 A. (0.30times 2times 22.7)/(24.31) B (0.30times 22.7)/(24.31) C. (0.30times 24.31)/(22.7) D (0.30times 22.7)/(24.31times 2)

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The correct answer is A. Explanation:1. The balanced chemical equation shows that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.2. The molar mass of magnesium is 24.31 g/mol.3. Given that 0.30 g of magnesium is reacting, we can calculate the number of moles of magnesium using the formula: moles = mass / molar mass.4. Moles of magnesium = 0.30 g / 24.31 g/mol = 0.0123 mol.5. Since 1 mole of magnesium produces 1 mole of hydrogen gas, 0.0123 mol of magnesium will produce 0.0123 mol of hydrogen gas.6. The volume of hydrogen gas produced can be calculated using the molar volume of an ideal gas at STP, which is 22.7 dm^3/mol.7. Volume of hydrogen gas = moles of hydrogen gas * molar volume of an ideal gas at STP = 0.0123 mol * 22.7 dm^3/mol = 0.2796 dm^3.8. Therefore, the correct expression for the volume of hydrogen gas produced at STP when 0.30 g of magnesium reacts with excess hydrochloric acid solution is , which corresponds to option A.