Question

MnO2+ 4HCL--->MnCl2+Cl2+2H2O:

if 0.86 mole of MnO2 and 48.2g of HCL react, which reagent will be used up first? How many grams of Cl2 will be produced?
The answers are HCL, and 23.4g.
Show the formula of these answers.

Answer 1

HCl is the reagent that will be used up first in the reaction between MnO2 and HCl. The reaction will produce 23.4 grams of Cl2 based on the stoichiometry of the balanced equation and the amount of the limiting reagent, HCl.

Step-by-step explanation:

To determine which reagent will be used up first in the reaction MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O, we need to perform a stoichiometric calculation.

First, calculate the moles of HCl present by using its molar mass (Molar mass of HCl = 36.46 g/mol):

Moles of HCl = 48.2 g / 36.46 g/mol = 1.32 moles

Given that the reaction requires 4 moles of HCl for each mole of MnO2, we can now compare the amount of HCl needed for the reaction with the actual amount provided:

Moles of HCl required for 0.86 moles of MnO2 = 0.86 moles MnO2 × 4 moles HCl/mole MnO2 = 3.44 moles HCl

Since we only have 1.32 moles of HCl available, HCl will be the limiting reagent and will be used up first.

To calculate the amount of Cl2 produced, we use the stoichiometry of the balanced equation, knowing that one mole of MnO2 produces one mole of Cl2:

Moles of Cl2 produced = Moles of HCl available / 4 = 1.32 moles / 4 = 0.33 moles Cl2

Using the molar mass of Cl2 (Molar mass of Cl2 = 70.90 g/mol) we find:

Mass of Cl2 produced = 0.33 moles × 70.90 g/mol = 23.4 grams of Cl2

Answer 2

Which reagent will be used up first? Thee answer is HCl.
We need to convert the amounts of the reactants into moles to see the limiting reactant.

48.2 g HCl ( 1 mol / 36.46 g) = 1.32 mol HCl

Since the ration is 1:4 then the limiting reactant is HCl. To react all of the MnO2 we need 3.44 mol HCl.

How many grams of Cl2 will be produced?
We use the amount of limiting reactant since it will be used up first.

1.32 mol HCl (1 mol Cl2 / 4 mol HCl) ( 70.9 g Cl2 / 1 mol Cl2) =23.4 g CL2