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10g of MnO(2)on reactoin with HCI forms ...

10g of` MnO_(2)`on reactoin with HCI forms 2.24 L of `Cl_(2)(g) at NTP_(1)` the percentage impurity of `MnO_(2) is :-
MnO_(2)+4HCl to MnCl_(2)+Cl_(2)+2H_(2)O`

A

`87%`

B

`25%`

C

`33.03%`

D

`13%`

Text Solution

Verified by Experts

The correct Answer is:
D
`MnO_(2)+4HCItoMnCI_(2)+CI_(2)+2H_(2)O^(n)CI_(2)=(2.24)/(22.4)=0.1"mole"`
`{:(,1 "mole"," ",1"mole"),(,87gm,,1 "mole"):}`
1 mole `CI_(2)` produced by 87 gm pure `MnO_(2)`
0.1 "mole" `CI_(2)` "produced by 8.7 gm pure" ` MnO_(2)`
`% "purity" = (8.7)/(10)xx100=87%`
% Impurity `= 13 %`
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