MnO2 reacts with HCI to form MnCI2 , CI2, and H2O determine the amount of each chemical present at the end of the reaction if 74.8 g of manganese iv oxide reacts with 48.2 grams of hydrochloric acid

Question

A. g of manganese iv oxide?

B. g hydrochloric acid?

C. g manganese ii chloride?

D. g chlorine?

E. g H2O

F. Verify the law of conservation of mass using your results

Answer 1

MnO₂ + 4HCl ==> MnCl₂ + Cl₂ + 2H₂O ... balanced equation

First, find moles of each reactant present:

moles MnO₂ = 74.8 g x 1 mole/86.94 g = 0.8604 moles

moles HCl = 48.2 g x 1 mole/36.5 g = 1.321 moles

Next, find which reactant is limiting:

There are several ways to do this. One of easiest is to simply divide the moles of each by the coefficient in the balanced equation and see which is less.

-For MnO₂ we have 0.8604 moles/1 = 0.8604

-For HCl we have 1.321 moles/4 = 0.3301 <= THIS IS THE LIMITING REACTANT

Next, using the limiting reactant, determine moles and mass of each product formed:

MnCl₂: 1.321 moles HCl x 1 mol MnCl₂/4 mol HCl x 125.8 g/mol = 41.5 g MnCl₂ (answer to C)

Cl₂: 1.321 moles HCl x 1 mol Cl₂/4 mol HCl x 70.9 g/mol = 23.4 g Cl₂ (answer to D)

H₂O: 1.321 moles HCl x 2 mol H₂O/4 mol HCl x 18 g/mol = 11.9 g H₂O (answer to E)

Finally, determine moles and mass of reactants that remain at the end of the reaction:

MnO₂: 1.321 moles HCl x 1 mol MnO₂ used/4 mol HCl = 0.330 moles used up. This leaves 0.8604 - 0.330 = 0.530 moles MnO₂ left over. 0.530 moles x 86.9 g/mol = 46.1 g MnO₂ left (answer to A)

HCl: Assuming the reaction runs to completion, there will be no HCl left over as it is limiting and will all be used up (answer to B)

F. To verify the law of conservation of mass, add up the masses before and after the reaction.

∑mass of before reaction = 74.8 g + 48.2 g = 123 g

∑mass after reaction = 41.5 g + 23.4 g + 11.9 g + 46.1 g = 122.9 g = 123 g