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Sagot :
Answer:
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.
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