Given the reaction `4HCl + MnO_2 -> MnCl_2 + 2H_2O + Cl_2` you have 41.3 g of `MnO_2` and 41.9 g of `HCl` . What is the theoretical yield of `Cl_2`

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The molar mass of `MnO_2` is 87 grams.  Applying a conversion factor we find `(41.3 g)/1 xx (1 mol)/(87 g)=.47 mol` of `MnO_2`

The molar mass of `HCl` is 36 grams.  Applying a conversion factor we find `(41.9 g)/1 xx (1 mol)/(36 g)=1.16 mol` of `HCl`

`.47/1.16 != 1/4`` `  Specifically `.47/1.16 > 1/4` which indicates that the HCl is the limiting reactant.  1.16 mol of HCl will react with only .29 mol of `MnO_2`

The ratio of HCl to `Cl_2` in the chemical equation is 4:1, to find the amount of `Cl_2` that should be produced, we apply equivalent ratios.  `1.16/n = 4/1` 4n=1.16 n=.29  .29 moles of `Cl_2` will be produced.  The Molar mass of `Cl_2` is 70 grams.  `(.29 mol)/1 xx (70 g)/(1 mol)=20.3g`

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