This is a limiting reagent (LR) problem. You know that when values are given for BOTH reactants.
Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)
mols HCl = M x L = approx 0.038 but you need a better answer than that estimate.
mols Mg(OH)2 = approx 0.05
Convert mols HCl to mols MgCl2 if we had all of the Mg(OH)2 we needed.
That's 0.038 x (1 mol MgCl2/2 mols HCl) = approx 0.038 x 1/2 = approx 0.019
Convert mols Mg(OH)2 to mols MgCl2 if we had all of the HCl needed. That's
0.05 x (1 mol MgCl2/1 mol Mg(OH)2) = 0.0 x 1/1 = approx 0.05
Both of these answers can't be right. In LR problems the correct value is ALWAYS the smaller value and the reagent producing that value is the LR. So HCl is the LR and approx 0.019 mols MgCl2 will be formed. I wouldn't pick A as the answer until going through and getting better numbers than my estimates.
How many moles of magnesium chloride are formed when 55 mL of 0.70 M HCl is added to 42 mL of 1.204 g Mg(OH)2?
Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)
A. 0.019 mol
B. 0.012 mol
C. 0.021 mol
D. 0.039 mol
1 answer