10 g of MnO2 on reaction with HCl forms 2.24 L of Cl2 at NTP . Then what is thwrong %impurity of MnCl2?
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MnO2 +4HCl --->MnCl2 +2H2O + Cl2
22.4 L of chlorine is produced from 87g of MnO2
hence 2.24 L of chlorine will be formed from 8.7 g of MnO2
hence impurity present in MnO2 = (10 - 8.7 ) g = 1.3 g
hence % age impurity of MnO2 = 1.3/10 x 100 = 13%
22.4 L of chlorine is produced from 87g of MnO2
hence 2.24 L of chlorine will be formed from 8.7 g of MnO2
hence impurity present in MnO2 = (10 - 8.7 ) g = 1.3 g
hence % age impurity of MnO2 = 1.3/10 x 100 = 13%
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