Chlorine is prepared in the laboratory by treating manganese dioxide `(MnO_(2))` with aqueous hydrochloric acid according to the reaction
`4HCl(aq)+MnO_(2)(s)to2H_(2)O(l)+MnCl_(2)(aq)+Cl_(2)(g)`
How many grams of HCl react with `5.0 g` of manganese dioxide ?

Explanation

Moles of `MnO_(2)=5/87=0.0574`
The reaction between `MnO_(2)` and HCl given is
`4HCl(aq)+MnO_(2)(s)to2H_(2)O(l)+MnCl_(2)(aq)+Cl_(2)(g)`
As per the above reaction for 1 mol. Of `MnO_(2)` 1 mol. Of `Cl_(2)` is produced by the reaction with 4 mol. of HCl.
`therefore` 1 mol. of `MnO_(2)` react with `4xx36.5` gmm HCl. 0.0574 mol. of `MnO_(2)` react with ? HCl
`(0.0574xx4xx36.5)/1=8.38` gm.