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

1 Answer
Apr 26, 2017

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