Answer:
a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂
b) Ni(OH)₂
c) KOH
d) 0.927 g
e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M
Explanation:
a) The equation is:
2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)
b) The precipitate formed is Ni(OH)₂
c) The limiting reactant is:
[tex] n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles [/tex]
[tex] n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles [/tex]
From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:
[tex] n = \frac{2}{1}*0.030 moles = 0.060 moles [/tex]
Hence, the limiting reactant is KOH.
d) The mass of the precipitate formed is:
[tex] n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles [/tex]
[tex] m = n*M = 0.010 moles*92.72 g/mol = 0.927 g [/tex]
e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:
[tex]C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M[/tex]
[tex]C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M[/tex]
[tex]C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M[/tex]
I hope it helps you!
The mass of precipitate formed is 0.93 g .
The equation of the reaction is;
NiSO4(aq) + 2KOH(aq) ---->K2SO4(aq) + Ni(OH)2(s)
The precipitate that forms is Ni(OH)2(s).
To obtain the limiting reactant;
Number of moles of NiSO4 = 200/1000 × 0.150 M = 0.03 moles
Number of moles of KOH = 100/1000 × 0.200 M = 0.02 moles
Since 1 mole of NiSO4 reacts with 2 moles of KOH
x moles of NiSO4 reacts with 0.02 moles of KOH
x = 1 mole × 0.02 moles/2 moles = 0.01 moles of NiSO4
Hence, KOH is the limiting reactant.
Now;
Since 2 mole of KOH yields 1 mole of Ni(OH)2
0.02 moles of KOH yields 0.02 moles × 1 mole/ 2 mole = 0.01 moles of Ni(OH)2
Mass of Ni(OH)2 produced = 0.01 moles of Ni(OH)2 × 93 g/mol = 0.93 g
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