Question

What volume of 7.91M solution of nitric acid, HNO3 is just sufficient to react with 15.9 g of lead dioxide, PbO2 ? 2PbO2 + 4HNO3 ---> 2PbNO3 + 2H20 + O2

Answer 1

`1.68 * 10^(-2)L`

Explanation:

First, convert the grams of lead dioxide into moles.

`mols PbO_2 = (15.9 g)/(239.198 g/(mol)) = 0.066472128 mols`

Since for every mole of lead dioxide you need 2 moles of nitric acid, your total moles of needed nitric acid would be:

`mols HNO_3 = 0.132944255 mols`

Because `M = (mols)/L`, you can rearrange the formula to get:

`L = (mols)/M`

`"liters nitric acid" = (0.132944255 mols)/(7.91M)`

`"liters nitric acid" = 0.016807112L`

Seeing that you only have 3 sig figs, the answer would be:

`1.68 * 10^(-2)L` of a 7.91M solution of nitric acid