Chemical Reaction Stoichiometry: Calculating Grams of Cl2 Produced

How many grams of Cl2 can be prepared from the reaction of 15.6 g of MnO2 and 30.8 g of HCl?

Final answer:

To calculate the grams of Cl2 produced from the reaction of MnO2 and HCl, we need to find the limiting reactant and use stoichiometry. The limiting reactant is MnO2, and we can produce 12.71 grams of Cl2.

Answer:

The reaction of 15.6 g of MnO2 and 30.8 g of HCl will produce 12.71 grams of Cl2.

When determining the amount of Cl2 that can be produced from the given reactants, we must first identify the limiting reactant. In this case, the limiting reactant is MnO2 based on the stoichiometry of the reaction.

The balanced chemical equation for the reaction is:

2 MnO2(s) + 8 HCl(aq) → 2 MnCl2(aq) + 5 Cl2(g) + 4 H2O(l)

From the equation, we can see that 2 moles of MnO2 react with 8 moles of HCl to produce 5 moles of Cl2. This gives us a stoichiometric ratio of 2:5 between MnO2 and Cl2.

By calculating the number of moles of each reactant, we find that 15.6 g of MnO2 is equivalent to 0.179 mol of MnO2, while 30.8 g of HCl is equivalent to 0.845 mol of HCl. Since the stoichiometric ratio is 2:5, MnO2 is the limiting reactant as it would require more moles of HCl to react completely.

Therefore, the amount of Cl2 produced will be based on the amount of MnO2 used. By converting 0.179 mol of Cl2 to grams using the molar mass of Cl2 (70.90 g/mol), we find that the reaction will produce a total of 12.71 grams of Cl2.