Question

2.00 g of I2 was produced in the reaction: 2KMnO4 + 10 KI + 8 H2SO4 → 6 K2SO4 + 2 MnSO4 + 5 I2 + 8 H2O

How much of KMnO4 in grams is required if it is 95% pure?

Answer 1

The mass of KMnO4 required is 0.497 g. To find the mass of KMnO4 required, convert the given 2.00 g of I2 to moles, then use stoichiometry to relate the moles of I2 to the moles of KMnO4. Finally, convert the moles of KMnO4 to grams by multiplying by the molar mass of KMnO4.

Step-by-step explanation:

To determine the amount of KMnO4 required, we can use stoichiometry and the given information.

The balanced equation shows that 2 moles of KMnO4 produce 5 moles of I2.

So, using the given 2.00 g of I2 and the molar mass of I2 (253.8 g/mol), we can calculate the moles of I2 produced.

moles of I2 = mass of I2 / molar mass of I2

= 2.00 g / 253.8 g/mol

= 0.00788 mol

Since 2 moles of KMnO4 produce 5 moles of I2, the moles of KMnO4 required can be found using the ratio:

moles of KMnO4 = moles of I2 * (2 mol KMnO4 / 5 mol I2)

= 0.00788 mol * (2 mol KMnO4 / 5 mol I2)

= 0.00315 mol KMnO4

Finally, to find the mass of KMnO4, we can multiply the moles of KMnO4 by the molar mass of KMnO4:

mass of KMnO4 = moles of KMnO4 * molar mass of KMnO4

= 0.00315 mol * 158.04 g/mol

= 0.497 g