Final answer:
To calculate the volume of hydrogen gas produced from 6.0g of magnesium, you need to use stoichiometry and the molar ratio between magnesium and hydrogen gas. The volume of hydrogen gas produced is 5.52 L.
Step-by-step explanation:
To find the volume of hydrogen gas produced, we need to use stoichiometry and the molar ratio between magnesium and hydrogen gas. The balanced chemical equation for the reaction is: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). From the equation, we can see that 1 mole of magnesium reacts with 2 moles of HCl to produce 1 mole of hydrogen gas. To calculate the volume of hydrogen gas produced, we need to convert the mass of magnesium to moles using its molar mass (24.305 g/mol) and then use the mole ratio to find the moles of hydrogen gas produced. Finally, we can use the ideal gas law to calculate the volume of hydrogen gas at standard temperature and pressure (STP).
Given: Mass of magnesium = 6.0 g. Molar mass of magnesium = 24.305 g/mol. Molar ratio between magnesium and hydrogen gas: 1:1. Molar volume at STP = 22.4 L/mol.
First, calculate the moles of magnesium: 6.0 g Mg × (1 mol Mg / 24.305 g Mg) = 0.2468 mol Mg.
Since the mole ratio between magnesium and hydrogen gas is 1:1, 0.2468 mol of magnesium will produce the same number of moles of hydrogen gas.
Finally, use the ideal gas law to calculate the volume of hydrogen gas at STP: Volume of H₂ = 0.2468 mol H₂ × 22.4 L/mol = 5.52 L.