2HCl + Mg(OH)2 ---> 2H2O + MgCl2

Question

_{a. If a reaction was conducted with 3.5 mol of HCl and 3.5 mol of M(OH)2, how much (in mol) magnesium chloride could be formed? explain why}

_{b. If a reaction was conducted with 3.5 g of HCl and 3.5 g og Mg(OH)2, how much (in g) magnesium chloride could be formed? explain why}

_{c. For question b, if 2.6 of MgCl2 was obtained after the completion of the chemical reaction. What is the % yield of the reaction?}

Answer 1

2HCl + Mg(OH)₂ ---> 2H₂O + MgCl₂ ... balanced equation

(a). Limiting reactant would be HCl since it takes twice as much HCl as Mg(OH)₂ according to the balanced equations. Thus, the 3.5 mols of HCl will determine how much MgCl₂ can be made.

3.5 mols HCl x 1 mol MgCl₂ / 2 mols HCl = 1.75 mols MgCl₂ can be formed

(b). For this question, we need to convert to moles from the grams.

Molar mass HCl = 36.5 g/mol

Molar mass Mg(OH)₂ = 58.3 g/mol

mols HCl = 3.5 g HCl x 1 mol / 36.5 g = 0.0959 mols HCl

mols Mg(OH)₂ = 3.5 g Mg(OH)2 x 1 mol / 58.3 g = 0.0600 mols Mg(OH)₂

Once again, HCl is limiting because there isn't twice as much HCl as Mg(OH)₂ which would be required according to the balanced equation.

Mass MgCl₂ formed = 0.0959 mols HCl x 1 mol MgCl₂ / 2 mol HCl x 95.2 g MgCl₂/mol = 4.6 g MgCl₂

(c). % yield = actual yield / theoretical yield (x100%) = 2.6 g / 4.6 g (x100%) = 57% yield