magnesium metal (1.00 mol) and a volume of aqueous hydrochloric acid that contains .500 mol of HCL are combined and react to completion. how many liters of hydrogen gas, measured at STP, are produced?

Mg(s) + 2HCL(aq) -> MgCl2(aq) + H2(g)

This is a limiting regent (LR) problem because amounts are given for BOTH reactants.

Mg + 2HCl ==> MgCl2 + H2
mols Mg = 1.00 mol
mol HCl = 0.5M

mols H2 produced from 1. mol Mg + 1 mol
mols H2 produced from 0.5 mol HCl = 1 mol
So this isn't a LR problem after all but the reactants just match. So 1 mol H2 is produced and at STP 1 mol occupies 22.4 L.

To answer this question, we need to use stoichiometry to determine the number of moles of hydrogen gas produced, and then convert it to liters at standard temperature and pressure (STP).

1. Start by determining the limiting reactant. This is the reactant that is completely consumed and determines the maximum amount of product that can be formed. In this case, we have 1.00 mol of magnesium metal and 0.500 mol of hydrochloric acid. The stoichiometry of the balanced equation tells us that 1 mol of magnesium reacts with 2 mol of hydrochloric acid to produce 1 mol of hydrogen gas. Therefore, 1.00 mol of magnesium requires 2.00 mol of hydrochloric acid to react completely. Since we have only 0.500 mol of hydrochloric acid, it is the limiting reactant.

2. Now let's calculate the moles of hydrogen gas produced. From the balanced equation, we know that 1 mol of magnesium reacts to produce 1 mol of hydrogen gas. Since hydrochloric acid is the limiting reactant, we can say that 0.500 mol of hydrogen gas is produced.

3. To convert moles of hydrogen gas to liters at STP, we need to use the ideal gas law, which states that PV = nRT. At STP, the temperature (T) is 273.15 K and the pressure (P) is 1 atm. The gas constant (R) is 0.0821 L·atm/(mol·K). Rearranging the equation to solve for volume (V), we get V = nRT/P.

4. Plugging in the values, we have n = 0.500 mol, R = 0.0821 L·atm/(mol·K), T = 273.15 K, and P = 1 atm. Substituting these values into the equation, we get V = (0.500 mol)(0.0821 L·atm/(mol·K))(273.15 K)/(1 atm) = 11.27 L.

Therefore, when 1.00 mol of magnesium metal and 0.500 mol of hydrochloric acid react to completion, 11.27 liters of hydrogen gas, measured at STP, are produced.