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A solution of 100 mL of 0.200 M KOH is mixed with a solution of 200 mL of 0.150 M NiSO4
(a) What precipitate forms?
(b) What is the limiting reactant?
(c) How many grams of the precipitate is formed?
(d) What is the concentration of each ion that remains in solution?
S7P6A66 (interaction - chemical).
S7P1A15 (containership - mass).
Pj Problem of Interest is of type interaction (chemical).
(a) Chemical equation:
2KOH + NiSO4 ------> K2SO4 + Ni(OH)2
Precipitate is Ni(OH)2
(b) Mol of KOH = (0.100)(0.200) = 0.0200 mol
Mol of NiSO4 =(0.200)(0.150) = 0.03 mol
Mole ratio of KOH:NiSO4 in balanced equation is 2:1
Mole ratio of KOH:NiSO4 in solution is 1:1.5
So, KOH is the limiting reactant.
(c) 2 moles of KOH forms 1 mole of Ni(OH)2
So, 0.0200 mol of KOH forms 0.0200/2 = 0.01 mol of Ni(OH)2
Molar mass of Ni(OH)2 = 59 + 32 + 2 = 93 g
So, gram of precipitate = 93(0.01) = 0.93 g of Ni(OH)2.
(d) From calculations in (b):
Initial mol of KOH = 0.02 mol
So, inital mol of K
So, initial mol of OH- = 0.02
Initial mol of NiSO4 = 0.03 mol
So, initial mol of Ni2+ = 0.03
So, initial mol of SO42- = 0.03
Mole of Ni(OH)2 in product = 0.01
So mol of N2+ in product = 0.01
So, mol of 2OH- = 0.02
So, mol of Ni2+ that remain in solution = 0.03 - 0.01 = 0.02 mol
Combined volume of 0.100 L solution of KOH and 0.200 L solution of NiSO4 = 0.3 L
So, concentration of Ni2+ that remains in solution = 0.02/0.3 = 0.0667 M.
Initial mole of K+ remains in solution:
So, concentration of K+ that remains in solution = 0.02/0.3 = 0.0667 M.
Similarly, initial mol of SO42- remains in solution
So, concentraion of SO42- that remains in solution = 0.03/0.3 = 0.1 M.
The point . is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.
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