Question

Calculate volume of o2 formed at STP when 2 moles of MnO2, react with 2 moles of H2O2 in excess of H2SO4, as per following reaction
MNO2+H2O2--->MNSO4+2H2O+O2​

Answer 1

Explanation:

2h²O² = 2h2o2

2 mol 1 mol

= 2× 34=68 g 22400ml

Answer 2

Answer:

44.8

Explanation:

Since H2O2 is the limiting reagent,

2 mole = volume of H2O / 22.4

Therefore, volume of H2O = 2 * 22.4

= 44.8L