Given the following unbalanced equation, how much CI2 is produced if 15.0 g MnO2 is reacted with 30.0 g HCI?

MnO2 + HCI----------MnCI2 + CI2 + H2O

The balanced equation is:
MnO2 + 4HCl ------> MnCl2 + Cl2 + 2H20

Next calculate the limiting reagent. Basically, the reactant that has lowest mole ratio.

molar mass MnO2 = 54.94g/mol + 2(16.00 g/mol) = 86.94
molar mass HCl = 1.008 + 35.45 = 36.46 g/mol

moles MnO2 = 15.0g/86.94 g/mol = 0.173 mol
moles HCl = 30.0g/36.46 g/mol = 0.823 mol

mole ratio MnO2 = 0.173 mol/1 mol = 0.173
mole ratio HCl = 0.823mol/ 4 mol = 0.206

The limiting reagent is MnO2.

One mole of Cl2 is produced for every one mole of MnO2 produced. 0.173 moles of MnO2 reacted, so 0.173 moles of Cl2 were produced.
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