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

1.68102L

Explanation:

First, convert the grams of lead dioxide into moles.

molsPbO2=15.9g239.198gmol=0.066472128mols

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

molsHNO3=0.132944255mols

Because M=molsL, you can rearrange the formula to get:

L=molsM

liters nitric acid=0.132944255mols7.91M

liters nitric acid=0.016807112L

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

1.68102L of a 7.91M solution of nitric acid