How many grams of the preciptate are formed? Chemistry help needed!!!!?

Question

A solution of 358 mL of 1.95 M KOH is mixed with a solution of 244 mL of 0.471 M NiSO4. How many grams of the precipitate form?

Answer 1

Step 1.) Write a balanced reaction.
2KOH + NiSO4 -> K2SO4 + Ni(OH)2

Step 2.) Find out how many moles of each reactant you have:
Molar mass of KOH: 56.1
Molar mass of NiSO4: 262.85

358 mL of 1.95 M KOH = .358L * 1.95 mol/L = 0.6981 mol
244 mL of 0.471 M NiSO4 = .244L * 0.471 mol/L = 0.114924 mol

NiSO4 is the limiting reactant. If .115 moles of NiSO4 react, then .230 moles of KOH react, and .115 moles each of K2SO4 and Ni(OH)2 are created.

K2SO4 is a soluble salt, but Ni(OH)2 is insoluble, so it will be the precipitate.

Step 3: Calculate the mass of the precipitate.
.115 moles of Ni(OH)2 * 92.7 g/mol = 10.7 grams Ni(OH)2 (solution!)