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)

The correct answer is A. $\frac {0.30\times 2\times 22.7}{24.31}$<br /><br />Explanation:<br />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.<br />2. The molar mass of magnesium is 24.31 g/mol.<br />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.<br />4. Moles of magnesium = 0.30 g / 24.31 g/mol = 0.0123 mol.<br />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.<br />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.<br />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.<br />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 $\frac {0.30\times 2\times 22.7}{24.31}$, which corresponds to option A.