HCl of specific gravity 1.2 g/mL and 5% by weight needed to produce 1.12L of Cl2 at STP by the reaction (as nearest integer). MnO2+4HCl→MnCl2+3H2O+Cl2
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Solution
(a) NHCl=%byweight×10×dEwofHCl =5×10×1.236.5=1.64 Now, mEq of MnO2= mEq of HCl =mEq of Cl2 formed. =1.1211.2×103 =100[EwofCl2=M21eqofCl2=11.2L]
(b) Volume of HCl used: N×V=100 1.64×V=100 ∴VHCl=60.97mL Because HCl is also used to give MnCl2 thus volume used is double than required for the reduction of MnO2. ∴VHCl=2×60.97=121.94mL
(c) Also, mEq of MnO2=mEq of HCl=100 W87/2×103=100(EwofMnO2=55+322) ∴WMnO2=4.35g nearest integer value is 4gm.