4HCl(aq) + MnO2(s) - 2H2O(L) + MnCl2(aq) + Cl2(g).
How many grams of HCl would react with 5 grams on MnO2, if Mn=55u and Cl=35.5u ?

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Solution

4HCl(aq) + MnO₂(s) $rightwards arrow$ 2H₂O(l) + MnCl₂(aq) + Cl₂(g).
number of moles 4 1 2 1 1

molecular mass of HCl = 1 + 35.5 = 36.5g/mol
molecular mass of MnO₂ = 55 + 2x 16 = 87g/mol

1 mole of MnO₂reacts with = 4 moles of HCl
ie 87 g of MnO₂reacts with = (4x36.5) g of HCl
1g of MnO₂reacts with = (4x36.5)/87 g of HCl
5g of of MnO₂reacts with = (4x36.5x5)/87 g of HCl
= 8.39 g of HCl

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