The weight of MnO2 and the volume of HCl of specific gravity 1.2 g/mL and 4% nature by weight needed to produce 1.78 L of Cl2 at STP. The reaction involved is as follows: MnO2+4HCl→MnCl2+2H2O+Cl2
A
True
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B
False
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Solution
The correct option is B False MnO2+4HCl→MnCl2+2H2O+Cl2 Moles 0.084×0.81.7822.4=0.08 =0.32 ⇒WMnO2=0.08×87g=6.96g; (10×4×1.236.5)×VL=0.32⇒VL=0.24L