Stoichiometric question .... confusing me... The answer would be a) 4cc b) 6.53 gm.

Question

5 ml of 8 N HNO3, 4.8 ml of 5 N HCl and a certain volume of 17 M H2SO4 are mixed together and made upto 2L. 30 ml of the acid mixture exactly neutralizes 42.9 ml of Na2CO3 solution containing 0.1 gm of Na2CO3.10H2O in 10 ml water . Calculate:

A) The volume of H2SO4 added to the mixture

B) The amount of the sulphate ions in the solution

Answer 1

ffirst let's find the grams of Na2CO3.10H2O  used:

42.9ml @ 0.1 g / 10 ml = 0.429 grams of Na2CO3.10H2O  were used

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next let's find moles of Na2CO3, using its molar mass of 286.14 g/mol:

0.429 g Na2CO3 @ 286.14g/mol = 1.50e-3 moles of Na2CO3,10H2O

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now let's find the moles of H+ ion that reacted with those moles of Na2CO3:

if:    2 moles H+ ion & 1 mole Na2CO3 --> salt, water & CO2

then

1.50e-3 mol Na2CO3.10H2O @ 2 moles H+ / 1 mole Na2CO3,10H2O = 3.00e-3 mol H+ ions

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now lets find the moles of H+ from the various sources:

HNO3: 0.005 litres @ 8 moles/litre = 0.04 Moles HNO3 total

? moles in 30 mls used @ 0.04 mol/2000 ml = 6.0e-4 moles H+

HCl: 0.0048 litres @ 5 mol/litre = 0.024 moles HCl total

? moles in 30 mls used @ 0.024 mol/2000 ml = 3.6e-4 moles H+

combining the two sources gives  9.6e-4 moles of H+ ion came from HCl & HNO3

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so, let's find the moles of H+ that came from the H2SO4:

3.00e-3 mol H+ ions total - 9.6e-4 moles  = 2.04e-3 moles of H+ must have come from the H2SO4

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let's find the moles of H2SO4 that released those H+ ions:

2.04e-3 mol H+ @ 1 mol H2SO4 / 2 mol H+ = 1.02 e-3 moles of  H2SO4 must have been added in the 30 mls

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so how many moles of H2SO4 was in the 2 litres of acid mix:

2000 ml @ 1.02e-3 mol / 30ml = 0.068 moles of H2SO4 must have been in the 2 litres

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but since those moles came from a 17Molar H2SO4 solution, how many ml of 17 Molar H2SO4 provided those moles of H2SO4:

0.068 Moles H2SO4 @ 17 moles / litre = 4.0e-3 litres of 17 molar H2SO4

that'syour first answer:  4 millilitres of 17 Molar H2SO4

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now let's get the last answer

B)The amount of the sulphate ions in the solution

if you have 0.068 moles of H2SO4 , the you have 0.068 moles of SO4-2 in the 2 litre solution

0.068 moles SO4-2 & a molar mass of 96 g/mol SO4-2 = 6.528 grams of sulfate are in the 2 liter solution

final answer = 6.53 g SO4-2