calculate the amount of grams in 0.25 mole of feSO4.7H2O
Answers
Answer:
(i)
Let x be the number of gram moles of hydrated ferric sulphate in mixture.
The molar mass of hydrated ferric sulphate Fe
2
(SO
4
)
3
.9H
2
O is 562 g/mol. The mass of hydrated ferric sulphate will be 562x grams.
Total mass of mixture is 5.5 g.
The mass of hydrated ferrous sulphate will be 5.5−562x grams. The molar mass of hydrated ferrous sulphate is 278 g/mol. The number of moles of hydrated ferrous sulphate will be
278
5.5−562x
This is equal to the number of moles of ferrous ions that can be oxidised by potassium permanganate.
5.4 mL of 0.1 N potassium permanganate are used.
Hence, number of g eq of potassium permanganate
=0.1geq/L×
1000ml/L
5.4ml
=0.00054 geq. They are also equal to the number of g eq of ferrous ions. They are also equal to number of moles of ferrous ions.
278
5.5−562x
=0.00054
5.5−562x=0.15012
562x=5.349
x=9.52×10
−3
moles =9.52 mmol.
Because 1 mole = 1000 mmol.
(ii) Let 100 g of mixture contains x moles of NO
2
The molecular weight of NO
2
is 46 g/mol. The mass of x moles will be 46 x grams.
The mixture will contains 100−46x grams of N
2
O
4
N
2
O
4
has molecular weigh of 92 g/mol. The number of moles of N
2
O
4
=
92g/mol
100−46x
The average molecular mass of the mixture is
x+
92
100−46x
46x+92(
92
100−46x
)
=
x+
92
100−46x
100
=
92x+100−46x
9200
=
46x+100
9200
1)
The vapour density of the mixture is 38.3. Its molecular weight is twice its vapour density. It is 2×38.3=76.6 g/mol (2)
(1) = (2)
46x+100
9200
=76.6
120.1=46x+100
46x=20.1
x=0.437 mol.
The number of moles of NO
2
is 0.437.