Question

When 200.0 mL of a 0.125 M sodium hydroxide is combined with 400.0 mL of a 0.025 M solution of magnesium nitrate, what is the percent yield if 0.45 g of precipitate is isolated?

Answer 1

Step-by-step explanation:

Balanced equation :

2NaOH (aq) + Mg(NO3)2 (aq) → Mg(OH)2(s) + 2NaNO3(aq)

2 mole NaOH reacts with 1 mole Mg(NO3)2 to produce 1 mole Mg(OH)2.

Check for limiting reactant :

Mol NaOH in 200.0 mL of 0.125 M solution

Mol = 0.125 mol/L * 200 mL / 1000 mL/L = 0.025 mol

This will react with 0.025 / 2 = 0.0125 mol Mg(NO3)2

Mol Mg(NO3)2 in 400 mL of 0.025 M solution

= 400 mL / 1000 mL/L * 0.025 mol /L = 0.01 mol Mg(NO3)2

The NaOH is in excess and the limiting reactant is Mg(NO3)2

0.01 mol Mg(NO3)2 will produce 0.01 mol Mg(OH)2

Molar mass Mg(OH)2 = 58.32 g/mol

Mass of 0.01 mol Mg(OH)2 = 0.01mol * 58.32 g/mol = 0.5832 g

Product isolated was 0.45 g

Percent yield = Product / Reactant * 100%

= 0.45 / 0.5832 * 100%

= 77.16% yield.