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

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

true

hope that

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g