Question

Chlorine is prepared in lab by treating MnO2 with aqueous HCl according to given reaction MnO2(s) + 4HCl (aq) → MnCl2 (aq) + Cl2(g) + 2H2O(l).

How many gram of chlorine is produced when 0.24 mole of HCl and excess MnO2 is taken?

Answer 1

Answer:

Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction. 4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl. How many grams of HCl react with 5.0 g of manganese dioxide? Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.

8.39 grams of Hcl would react with 5 gm of MnO2.

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

for this reaction determine the mass of HCl that will react with 5 gm of MnO2.

Mass of

MnO2 = 54.94 + 2 ( 15.99 ) = 86.92 g of MnO2.

Mass of

Hcl = 1.01 + 35.45 = 36.46 g of Hcl.

= 4 mole of Hcl

1 mole of Mno2.

= ( 5.00) (4) ( 36.46) / 86.92

= 8.39 g of Hcl

Hence Proved!!