By starting with 0.5 moles of sodium peroxide how many moles of dioxygen gas can be obtained by dropping excess of acidified potassium permanganate solution on it?

Question

By starting with 0.5 moles of sodium peroxide how many moles of dioxygen gas can be obtained by dropping excess of acidified potassium permanganate solution on it? (A) 0.125 mole (B) 1 mole (C) 0.25 mole (D) 0.5 mole

Tardigrade Admin · Accepted Answer

2KMnO4+5Na2O2+8H2SO4 arrow K2SO4+2MnSO4+5Na2SO4+8H2O+5O2 text5 textmole textNa text2 textO text2 textO text2 text= text5 gives textO text2 text= text5 textmole So text0 text.5 textmol textNa text2 textO text2 give textO text2 text= text0 text.5 textmol

Q. By starting with 0.5 moles of sodium peroxide how many moles of dioxygen gas can be obtained by dropping excess of acidified potassium permanganate solution on it?

0.125 mole

1 mole

0.25 mole

0.5 mole

$2 K M n O_{4} + 5 N a_{2} O_{2} + 8 H_{2} S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 5 N a_{2} S O_{4} + 8 H_{2} O + 5 O_{2}$
$5 mole Na_{2} O_{2} O_{2} = 5$ gives $O_{2} = 5 mole$
So $0 .5 mol Na_{2} O_{2}$ give $O_{2} = 0 .5 mol$

Q. By starting with 0.5 moles of sodium peroxide how many moles of dioxygen gas can be obtained by dropping excess of acidified potassium permanganate solution on it?

0.125 mole

1 mole

0.25 mole

0.5 mole

2KMnO4​+5Na2​O2​+8H2​SO4​→K2​SO4​+2MnSO4​+5Na2​SO4​+8H2​O+5O2​ 5moleNa2​O2​O2​=5 gives O2​=5mole So 0.5molNa2​O2​ give O2​=0.5mol

$2 K M n O_{4} + 5 N a_{2} O_{2} + 8 H_{2} S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 5 N a_{2} S O_{4} + 8 H_{2} O + 5 O_{2}$
$5 mole Na_{2} O_{2} O_{2} = 5$ gives $O_{2} = 5 mole$
So $0 .5 mol Na_{2} O_{2}$ give $O_{2} = 0 .5 mol$