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andre [41]
3 years ago
10

A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ

ation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?
Chemistry
1 answer:
NISA [10]3 years ago
5 0

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:

n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles

n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

n = \frac{2}{1}*0.030 moles = 0.060 moles                  

Hence, the limiting reactant is KOH.  

d) The mass of the precipitate formed is:

n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles

m = n*M = 0.010 moles*92.72 g/mol = 0.927 g  

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M  

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

C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M

I hope it helps you!                                                                        

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1. For every 5.00 mL of milk of magnesia there are 400. mg of magnesium hydroxide. How many mL of milk of magnesia do we need to
forsale [732]

Answer:

1. 6.50mL

2. pH = 13.097

Explanation:

1. The neutralization of HCl with Al(OH)₃ is:

3HCl + Al(OH)₃ → Al(Cl)₃ + 3H₂O

40.0mL ≡ 0.0400L of 0.500M HCl are:

0.0400L × (0.500mol / L) = <em><u>0.0200 moles of HCl</u></em>

Based on the neutralization, 3 moles of HCl react with 1 mole of Al(OH)₃, thus, the moles of Al(OH)₃ that react with 0.0200 mol of HCl are:

0.0200mol HCl × (1mol Al(OH)₃ / 3mol HCl) = <em><u>0.00667 moles of Al(OH)₃</u></em>. In grams:

0.00667 moles Al(OH)₃ × (78g / 1mol) = 0.520g of Al(OH)₃ ≡ 520mg

As 5.00mL of milk magnesia contain 400mg of Al(OH)₃, the mL of milk magnesia required for a complete reaction are:

520 mg Al(OH)₃ × (5.00mL / 400mg) =<em> 6.50mL</em>

<em />

2. The reaction of lithium hydroxide (LiOH) with perchloric acid (HClO₄) is:

HClO₄ + LiOH → LiClO₄ + H₂O

<em>The reaction is 1:1.</em>

Moles of LiOH and HClO₄ are:

LiOH: 0.0500L × (0.350mol / L) = <em>0.0175 moles of LiOH</em>

HClO₄: 0.0300L × (0.250mol / L) = <em>0.0075 moles of HClO₄</em>

Assuming the reaction goes to completion, moles of LiOH that remains are:

0.0175 mol - 0.0075 mol = 0.0100 moles of LiOH. The total volume is 80.0mL, 0.0800L. Thus, molarity of LiOH is:

0.0100 mol / 0.0800L = <em>0.125 M of LiOH</em>

It is possible to obtain the pOH of the solution thus:

pOH = -log (OH) = -log 0.125M = 0.903

As pH = 14- pOH,

pH = 14 - 0.903 = <em>13.097</em>

I hope it helps!

3 0
3 years ago
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