Question

Suppose you react 12.61 grams of MnO₂ according to the following (unbalanced) equation:

HCl + MnO₂→H₂O + MnCl₂ + Cl₂

₁₎How many moles of manganese(IV) oxide do you have?

2) How many moles of HCl do you need for a complete reaction?

3)How many grams of HCl do you need for a complete reaction?

4)How many grams of each product do you form?

5)Prove that mass has been conserved.

what is the total mass present before the reaction occurs?

What is the total mass present after the reaction occurs?

Answer 1

4HCl + MnO2 ---> 2H2O + MnCl2 + Cl2

4 mol HCl = 1 mol MnO2

1) No of mol of MnO2 = 12.61/87 = 0.145 mol

2) No of mol of HCl needed = 0.145*4/1 = 0.58 mol

3) mass of HCl needed for a complete reaction = 0.58*36.5 = 21.17 g

4) mass of water formed = 0.145*2*18 = 5.22 g

   mass of MnCl2 formed = 0.145*125.85 = 18.25 g

   mass of cl2 formed = 0.145*71 = 10.3 g

5)

total mass present before the reaction occurs

      = 12.61 +21.17 = 33.78 g

total mass present after the reaction occurs

       = 5.22+18.25+10.3

       = 33.77 g

as the total mass present before the reaction occurs = total mass present after the reaction occurs,

it obeys law of conservation of mass.