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Find the amound of iron pyrites `(FeS_2)` which is sufficient to produce enough `SO_2` on roasting (heating in excess of `O_2`) such that is `(SO_2)` completely decolourise a 1 L solution of `KMnO_4` containing 15.8 g `L^(-1)` of it. The equation are `FeS_2+O_2toFe_2O_3+SO_2` `KMnO_4+SO_2toMnSO_4+H_2SO_4+KHSO_4` |
Answer» First calculate the amount of `SO_2` required to decolourise `15.8gL^(-1)` of `KMnO_4` solution. For this, balance the following chemical reaction. The balanced equation is as: `KMnO_4+SO_2toMnSO_4+H_2SO_4+KHSO_4` `2KMnO_4+5SO_2+2H_2Oto2MnSO_4+H_2SO_4+2KHSO_4` `2" mol of "KMnO_4-=5 " mol of "SO_2` Calculate moles in `15.8g L^(-1) of KMnO_4` Using strength `(gL^(-1))=(M)/(Mw)` `implies1.0L of KMnO_4` contains 0.1 mol Hence, moles of `SO_4` required `=(5)/(2)(0.1)=0.25` To calculate the amount of pyrites, we have to balance the following reaction. `FeS_2+O_2toFe_2O_3+SO_2` Balancing the reaction, we have `4FeS_2+11O_2to2Fe_2O_3+8SO_2` From stoichiometry of roasting, we have: `8 " mol of "SO_2-=4" mol of "FeS_2` `0.25 " mol of "SO_2-=(4)/(8)(0.25)" mol of "FeS_2` `=0.125 " mol of "FeS_2` Mass of `FeS_2=0.125xx120=15gL^(-1)` |
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