1.

Find out the volume of Cl_(2) at STP produced by the action of 100cm^(3) of 0.2 N HCl on excess of MnO_(2).

Answer»


Solution :`MnO_(2)+underset(4xx36.5g)(4HCl)rarrMnCl_(2)+2H_(2)O+underset(22400cm^(3)" at STP")(Cl_(2))`
`"HCl PRESENT in "100cm^(3)" of 0.2 N HCl"=(0.2)/(1000)xx"100g eq.= 0.02g eq."xx"36.5 g = 0.73 g"`


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