We drop 30 grams of magnesium into 770 mL of a 4 M HCl solution. What is the maximum volume of dry hydrogen that could be produced by this reaction at STP? Mg(s) + 2HCl(aq)→MgCl2(aq) + H2(g)
Answer in units of liters
To determine the maximum volume of dry hydrogen gas produced by the reaction, we need to use stoichiometry and then apply the ideal gas law. Here are the steps to calculate the maximum volume:
1. Determine the number of moles of magnesium (Mg) used:
Molecular weight of Mg = 24.31 g/mol
Moles of Mg = Mass of Mg / Molecular weight of Mg = 30 g / 24.31 g/mol
2. Determine the number of moles of hydrochloric acid (HCl) reacting:
From the reaction equation, we see that 1 mole of Mg reacts with 2 moles of HCl.
Moles of HCl = Moles of Mg * (2 moles of HCl / 1 mole of Mg)
3. Calculate the volume of hydrogen gas at STP:
We know that 1 mole of any gas at STP occupies 22.4 L.
Volume of H2 gas = Moles of H2 * 22.4 L/mol
Let's perform the calculations:
1. Moles of Mg = 30 g / 24.31 g/mol = 1.23 mol
2. Moles of HCl = 1.23 mol * (2 moles of HCl / 1 mole of Mg) = 2.46 mol
3. Volume of H2 gas = 2.46 mol * 22.4 L/mol = 55.1 L
Therefore, the maximum volume of dry hydrogen gas that could be produced in this reaction at STP is 55.1 liters.