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10g of` MnO_(2)`on reactoin with HCI forms 2024 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%` |
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Answer» Correct Answer - 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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