Question

Identify the limiting reactant when .86g of MnO2 and 48.2g of HCl react. How many grams of Cl2 will be produced? MnO2 + 4HCl —> MnCl2 + Cl2 + 2H2O

Answer 1

we have to transform the mass of the reactants into mols to do that we have to divide the mass of each component by its molar mass. (molar mass of MnO2= 89.6g/mol; molar mass of HCl= 36.4 g/mol)

`\begin{gathered} n=(m)/(Mm) \\ n_(MnO_2)=(0.86g)/(0.86.9/mol)=0.0099mol \\ n_(HCl)=(48.2g)/(36.45g/mol)=1.32mol \end{gathered}`

To find the limiting reactant we chose one of the reactants, MnO2 in this case, and calculate how many mols of the other reactant (HCl) would be consumed for that amount of MnO2