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Calculate the weight of MnO_2 required to produce 1.50 L of chlorine at 27^@C and 1.50 atm pressure according to the following equation: MnO_2 + 4HCIto MnCl_2 + 2H_2 O + Cl_2 |
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Answer» Solution :The chemical EQUATION INVOLVED is `MnO_2 + 4HCIto MnCl_2 + 2H_2 O + Cl_2` `54.9+2xx16 "" 22.4 L` `=86.9 "" at S.T.P.` This equation implies that 86.9 g of `MnO_2`will give 22.4 of CHLORINE at S.T.P. Since the given conditions are different from those at S.T.P., therefore we should first convert the given volume corresponding to S.T.P. Therefore, `P_1 = 1.50 " atm, " V_1 = 1.50 L, T_1 = 27 + 273 = 300 K` `P_2 = 1 " atm,"V_2 = ?"" T_2 = 273 K` (S.T.P.) According to the gas equation, `(P_1 V_1)/T_1 = (P_2 V_2)/T_2` or`V_2 = (P_1 V_1 T_2)/(T_1 P_2)= ( 1.50 x× 1.50 xx273)/(300xx1)=2.05L` Thus, 1.5 L of `Cl_2` in the given conditions is equivalent to 2.05 L at S.T.P. `:."" 22.4 " L of chlorine at S.T.P. is given by " MnO_2 = 86.9` g `:.2.05 " L of chlorine at S.T.P. will be given by "MnO_2= (86.89)/(22.4)xx 2.05 = 7.95` g |
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