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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#

First, change the lead dioxide grams into moles.

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

You require two moles of nitric acid for every mole of lead dioxide, so the total amount of nitric acid you would require 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#

Given that you only have three significant figures, the response is:

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

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Answer 2

Henry Antrim

To react with 15.9 g of lead dioxide (PbO2), you would need 0.0500 L of 7.91 M nitric acid (HNO3) solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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 ---&gt; 2PbNO3 + 2H20 + O2

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