Chlorine is prepared in the laboratory by treating manganese dioxide \((MnO_2)\) with aqueous hydrochloric acid according to the reaction
\(4 HCl (aq) + MnO_2(s) → 2H_2O (l) + MnCl_2(aq) + Cl_2 (g)\).
How many grams of HCl react with 5.0 g of manganese dioxide?
\(4 HCl (aq) + MnO_2(s) → 2H_2O (l) + MnCl_2(aq) + Cl_2 (g)\).
How many grams of HCl react with 5.0 g of manganese dioxide?
Solution and Explanation
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 = \(\frac {146\ g }{ 87 \ g} × 5.0\ g\) 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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