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Review the equation :-
MnO2 + 2NaCl + 3H2SO4 → MnSO4 + 2NaHSO4 + Cl2 + 2H2O
In order to get 4.48 litre of Cl2 at NTP what volume of 18M H2SO4 is needed :-

Asked by Atulcaald 15th May 2018, 12:58 PM
Answered by Expert
Answer:
 
MnO subscript 2 space plus space 2 NaCl space plus space 3 straight H subscript 2 SO subscript 4 space rightwards arrow space MnSO subscript 4 space plus space 2 NaHSO subscript 4 space plus space Cl subscript 2 plus space 2 straight H subscript 2 straight O

Moles space of space chlorine

space 22.4 space straight l space equals space 1 mole space

space 4.48 space straight l space equals fraction numerator 4.48 cross times 1 over denominator 22.4 end fraction

space space space space space space space space space space space equals space 0.2 space mole space of space space Cl subscript 2

From space reaction semicolon

1 space mole space of space Cl subscript 2 space identical to space 3 space moles space of space straight H subscript 2 SO subscript 4

0.2 space mole space of space Cl subscript 2 space fraction numerator 0.2 cross times 3 over denominator 1 end fraction space moles space of space straight H subscript 2 SO subscript 4

moles space of space straight H subscript 2 SO subscript 4 equals space 0.6 space mole

Molarity space of space straight H subscript 2 SO subscript 4 space equals 18 space straight M

Molarity space equals fraction numerator No. space of space moles space over denominator Volume end fraction

Volume space equals space fraction numerator space No. of space moles over denominator Molarity end fraction

space space space space space space space space space space space space equals space 0.033 space straight l

space space space space space space space space space space space space space equals space 33.33 space ml
Answered by Expert 18th May 2018, 7:25 PM
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