Question

1) If a solution containing 131.15 g of silver nitrate is allowed to react completely with a solution containing 11.29 g of lithium hydroxide, how many grams of solid precipitate will be formed?

2) How many grams of the reactant in excess will remain after the reaction?

3) Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (0) for the number of moles.

Answer 1

1.

Write the reaction equation first:

AgNO₃(aq) + LiOH(aq) ------> LiNO₃ (aq) + AgOH(s)

169.87 g/mol 23.95 g/mol   124.88 g/mol

169.87 g silver nitrate give 124.88 g silver hydroxide

131.15g will give = 124.88 g / 169.87g x 131.15g = 96.413g

23.95g LiOH forms 124.88g AgOH

11.29g will form = 124.88g / 23.95g x 11.29g = 58.868g AgOH

limiting reactant is LiOH as it is giving lesser product and the amount will be based on limiting reactant

solid precipitate = 58.868g AgOH

2.

23.95 g LiOH need 169.87g AgNO₃ for complete reaction

11.29 g will need = 169.87g/23.95 g x 11.29g = 80.076g AgNO₃

excess reactant remaining = 131.15g - 80.076g(reacted) = 51.073g

3.

moles of LiOH reacted = moles of LiNO₃ formed = 11.29g/ 23.95 g/mol = 0.4714 mol

It should form 0.4714 mol Li⁺(aq) and 0.4714 mol NO₃^-(aq)

and remaining AgNO₃ will also get dissociated

= 51.073g/169.87 g/mol = 0.301 mol

It will give 0.301 mol NO₃^-(aq) and 0.301Ag⁺(aq)

total NO₃^-(aq) = 0.301 mol + 0.4714 mol = 0.7720 mol NO₃^-(aq)

total Ag⁺(aq) = 0.301 mol

total Li⁺(aq) = 0.4714 mol

OH^-(aq) ions = 0 moles