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Question
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction
\[\ce{4HCl_{(aq)} + MnO_{2(s)} → 2H2O_{(l)} + MnCl_{2(aq)} + Cl_{2(g)}}\]
How many grams of HCl react with 5.0 g of manganese dioxide?
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Solution
1 mol [55 + 2 × 16 = 87 g] MnO2 reacts completely with 4 mol [4 × 36.5 = 146 g] of HCl.
∴ 5.0 g of MnO2 will react with
`=(146 "g")/(87 "g")xx5.0 "g"` of HCl
= 8.4 g of HCl
Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.
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