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What volume of air at N.T.P containing 21 % oxygen by volume is required to completely burn 1000 g of sulphur containing 4 % incombustible matter ? |
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Answer» Correct Answer - `3360 L` Amount of incombustible matter in sulphur `= (1000 xx 4)/(100) = 40 g` Amount of pure matter `= 1000 - 40 = 960 g` Step I. Volume of oxygen required The chemical equation for the reaction is `underset(32g)(1//8S_(B))+underset(22.4L)(O_(2))rarrSO_(2)` `32g` of sulphur `(S_(8))` require `O_(2)` at `N.T.P.` `= 22.4 L` `960g` of sulphur `(S_(8))` require `O_(2)` at `N.T.P = ((22.4 L) xx (960g))/((32g)) = 972 L` Step II. Volume air required Volume of oxygen required `= 972 L` Volume of air required `= (672 L) xx (100)/(200) = 3360 L` |
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