checking some answers...thanks?

Question

7. Given the following equations and DHo values, determine the heat of reaction (kJ) at 298 K for the reaction:

PCl3(g) + Cl2(g) à PCl5(g)

2 P(s) + 3 Cl2(g) à 2 PCl3(g) DHo/kJ = -542
2 P(s) + 5 Cl2(g) à 2 PCl5(g) DHo/kJ = -764



8. 4.90g of of ammonium iron (II)sulfate hexahydrate, FeSO4(NH4)2SO4*6H2O, was dissolved in 200.mL of acidified water. 20.0mL of the solution reacted completely with 12.6mL of potassium manganate (VII) (KMnO4) solution according to the balanced equation below. Calculate the concentration of the potassium manganate (VII) solution

5Fe2+ + MnO4- + 8H+ à 5Fe3+ + Mn2+ + 4H2O

Answer 1

Problem 7.
2 P(s) + 3 Cl2(g) ==> 2 PCl3(g) DHo/kJ = -542...(1)
2 P(s) + 5 Cl2(g) ==> 2 PCl5(g) DHo/kJ = -764...(2)
{(2) - (1)}/2:
PCl3(g) + Cl2(g) ==> PCl5(g), DHo/kJ = -111

Problem 8.
First check to see whether the reaction is balenced:
5Fe(2+) + MnO4- + 8H+ ==> 5Fe(3+) + Mn(2+) + 4H2O
molar mass of FeSO4(NH4)2SO4*6H2O is: 392.14 g/mol.
Hence the concentration of the potassium manganate (VII) solution is:
4.90/(392.14*10*5)/0.0126 = 0.0198 (M)

Answer 2

(A) 2 P(s) + 3 Cl2(g) à 2 PCl3(g) DHo/kJ = -542
(B) 2 P(s) + 5 Cl2(g) à 2 PCl5(g) DHo/kJ = -764

1/2(B) - 1/2 (A) gives:

PCl3(g) + Cl2(g) à PCl5(g)
------
so 1/2(-764) - 1/2(-542)

gives: ANSWER = -111 kJ
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8. 4.90g of of ammonium iron (II)sulfate hexahydrate, FeSO4(NH4)2SO4*6H2O=

4.90 grams @ 392.3 g/mol = 0.0125 moles

but they only used 1/10 of it when they used 20 mls out of the 200 mls it was dissolved in=

0.00125 moles of hydrate
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0.00125 moles of hydrate has 0.00125 moles of Iron

0.00125 mol Fe reacts @ 1 mol MnO4 / 5 mol Fe =

0.000250 moles of MnO4 reacted
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0.000250 mol of MnO4 / 0.0126 litres =

0.0198 Molar