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Maslowich
3 years ago
12

Manganese(II) oxide, lead(IV) oxide, and nitric acid react to produce permanganic acid, lead(II) nitrate, and water according to

the reaction above. How many moles of each product could be produced by the reaction of 8.90 moles of nitric acid with excess manganese(II) oxide and excess lead(IV) oxide
Chemistry
1 answer:
Leokris [45]3 years ago
7 0

Answer:

There will be produced:

2.97 moles HMnO4

4.45 moles Pb(NO3)2

2.97 moles H2O

Explanation:

Step 1: Data given

Manganese(II) oxide = MnO2

lead(IV) oxide = PbO2

nitric acid = HNO3

Moles of HNO3 = 8.90 moles

Step 2: The balanced equation

2MnO2 + 3PbO2 + 6HNO3 → 2HMnO4 + 3Pb(NO3)2 + 2H2O

Step 3: Calculate moles of reactants and products

For 2 moles MnO2 we need 3 moles PbO2 and 6 moles HNO3 to produce 2 moles HMnO4, 3 moles Pb(NO3)2 and 2 moles of water

For 8.90 moles of HNO3, there will react:

8.90 / 3 = 2.97 moles MnO2

8.90 / 2 = 4.45 moles PbO2

There will be produced:

8.90/3 = 2.97 moles HMnO4

8.90/2 = 4.45 moles Pb(NO3)2

8.90 / 3 = 2.97 moles H2O

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(P_{1}) represents the initial pressure of the gas

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