# 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

Apr 26, 2017

$1.68 \cdot {10}^{- 2} L$

#### Explanation:

First, convert the grams of lead dioxide into moles.

$m o l s P b {O}_{2} = \frac{15.9 g}{239.198 \frac{g}{m o l}} = 0.066472128 m o l s$

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

$m o l s H N {O}_{3} = 0.132944255 m o l s$

Because $M = \frac{m o l s}{L}$, you can rearrange the formula to get:

$L = \frac{m o l s}{M}$

$\text{liters nitric acid} = \frac{0.132944255 m o l s}{7.91 M}$

$\text{liters nitric acid} = 0.016807112 L$

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

$1.68 \cdot {10}^{- 2} L$ of a 7.91M solution of nitric acid