Solved problem

Potassium permanganate (KMnO₄) oxidizes hydrogen peroxide (H₂O₂) to oxygen. What volume of 0.945

H₂O₂ would consume 125

of 0.0540

KMnO₄?

2MnO₄^–(aq) + 3H₂O₂(aq) → 2MnO₂(s) + 3O₂(g) + 2OH^–(aq) + 2H₂O

Known

c(MnO₄^–)
V(MnO₄^–)

n = cV
→
Step 1

n(MnO₄^–)

n(MnO₄^–)	=	n(H₂O₂)
2	=	3

→
Step 2

n(H₂O₂)

c	=	n
		V

→
Step 3

V(H₂O₂)

Unknown

n(MnO₄^–) = 125 × 10^–3

× 0.0540

n(MnO₄^–) = 6.75 × 10^–3 mol

6.75 × 10^–3 mol	=	n(H₂O₂)
2	=	3

n(H₂O₂) = 1.0125 x 10^–2 mol

0.945	=	1.0125 x 10^–2 mol
		V(H₂O₂)

V(H₂O₂) = 0.0107

= 10.7

Step 1: Use given information (concentration and volume) to calculate n(known).

Step 2: Calculate n(unknown) from n(known) and the coefficients on these substances in the balanced equation for the reaction.

Step 3: Use calculated n(unknown) and given c(unknown) to calculate V(unknown).

Note that intermediate values have more than the three significant figures in the given data. Rounding should be done only at the very end of the calculation.