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

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Answer 1 · 2017-04-26T23:57:27

#1.68 * 10^(-2)L#

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