9. 500ml of 0.3M HCLO4 is combined with 500ml of 0.2M Ca(OH)2. Calculate the mol
ID: 1066018 • Letter: 9
Question
9. 500ml of 0.3M HCLO4 is combined with 500ml of 0.2M Ca(OH)2. Calculate the molarity of H+, CLO4-, Ca+2 and OH- ions after the reaction.
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C8H18(l) + 25/2 O2(g)8 CO2(g)+ 9H2O(g) Hrxn=-5074.1kj
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CaSO4 2H2)(s)CaSO4(s) + H2)(g)
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Explanation / Answer
9. 500ml of 0.3M HCLO4 is combined with 500ml of 0.2M Ca(OH)2. Calculate the molarity of H+, CLO4-, Ca+2 and OH- ions after the reaction.
V = 500 mL = 0.5 L; M = 0.3
mol of HClO4 = MV = 0.3*0.5 = 0.15 mol of HClO4
for base
V = 500 mL = 0.5 L; M = 0.3 --> MV = 0.10 mol of Ca(OH)2
limiting reactant is acid, since 2:1 ratio
so
0.15 mol of HCl react with 0.075 mol of Ca(OH)2 to form
2HClO4 + Ca(OH)2 = Ca(ClO4)2 + 2H2O
Vfinal = V1+V2 = 500+500 = 1000 mL = 1 L
[H+] = 0 since all racts
[ClO4-] = 0.15 / 1 = 0.15 M
[Ca+2] = 0.10 / 1 = 0.10 M
mol of OH left = 2*(0.10 ) - 0.15 = 0.05 mol
[OH-] = 0.05 /1 = 0.05 M
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