Question

Given the balanced equation 2 Mg(s) + 2 HCl(aq) → 2 MgCl2(aq) + H2(g), if 15 moles of Mg react with 10 moles of HCl, calculate A) the grams of MgCl2 produced and B) identify the limiting reactant.

Answer 1

To calculate the grams of MgCl2 produced, use stoichiometry and convert from moles of MgCl2 to grams. The moles of MgCl2 produced will be equal to the moles of Mg. To identify the limiting reactant, compare the moles of each reactant to the stoichiometric ratio in the balanced equation. HCl is the limiting reactant.

Step-by-step explanation:

In order to calculate the grams of MgCl2 produced, we need to use stoichiometry and convert from moles of MgCl2 to grams. The molar mass of MgCl2 is 95.21 g/mol. Since the balanced equation shows a 1:1 ratio between moles of MgCl2 and moles of Mg, the number of moles of MgCl2 produced will be equal to the moles of Mg. Therefore, 15 moles of Mg will produce 15 moles of MgCl2. Multiplying the number of moles of MgCl2 (15 moles) by the molar mass of MgCl2 (95.21 g/mol) gives us the grams of MgCl2 produced:

15 moles x 95.21 g/mol = 1428.15 g MgCl2

To identify the limiting reactant, we need to compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The balanced equation shows a 2:2 ratio between moles of Mg and moles of HCl. Therefore, for every 2 moles of Mg, we need 2 moles of HCl. Using the given quantities, we can calculate the moles of each reactant:

Moles of Mg = 15

Moles of HCl = 10

Since the moles of HCl (10 moles) are less than the stoichiometric ratio of 2 moles per 2 moles of Mg, HCl is the limiting reactant.