How many milliliters of a 0.204 M HI solution are needed to reduce 24.5 mL of...

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How many milliliters of a 0.204 M HI solution are needed toreduce 24.5 mL of a 0.371 M KMnO4 solution according to thefollowing equation: 10HI + 2KMnO4 + 3H2SO4 â†’ 5I2 + 2MnSO4 + K2SO4 +8H2O

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number of moles of KMNO4 = M* V
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.371 M * 0.0245 L
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 9.0895*10^-3 mol

From reaction,
2 mol of KMNO4 requires 10 mol of HI
so, number of moles of HI required = (10/2)* number of moles of KMNO4
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 5 * 9.0895*10^-3 mol
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.04545

number of moles of HI required = M*V
0.04545 = 0.204 * V
V=0.223 L
Â Â = 223 mL
Answer: 223 mL

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