What volume of 0.170 M KMnO4 would be required to oxidize 20.0 mL of 0.310 M KI in acidic solution

The products include I2 and Mn+2.
1. 12.5 mL
2. 21.56 mL
3. 3.00 mL
4. 7.29 mL

find moles
0.0200 L (0.310 mol/Litre) KI = 0.0062 moles of KI

by the reaction
2 KMnO4 & 10 KI --> 5 I2 & 2 Mn+2
0.0062 moles of KI reacts with 1/5 th as many moles of KMnO4 = 0.00124 moles KMnO4

find volume
0.00124 moles KMnO4 (1 litre / 0.170 Mol KMnO4) = 0.00729 litres

which is 7.29 ml

your answer is
"4." 7.29 mL