141k views
0 votes
4. 34.3 g Pb3O4 is dissolved in 500 ml of 4 M HNO3 then weight of residue is? (Pb = 207)

User Maricela
by
4.8k points

1 Answer

1 vote
Answer:

The weight of the residue = 45.1 g

Step-by-step explanation:

The equation of reaction is:


Pb_3O_4+4HNO_3\rightarrow PbO_2+2Pb(NO_3)_2+2H_2O

The residue will contain a mixture of PbO₂ and Pb(NO₃)₂

Molar mass of Pb₃O₄ = (207 x 3) +(16 x 4) = 685 g/mol

Number of moles of Pb₃O₄ = Mass/Molar mass

Number of moles of Pb₃O₄ = 34.3/685

Number of moles of Pb₃O₄ = 0.05 moles

Number of moles of HNO₃ = 4 x 0.05 = 0.2 moles

Molar mass of HNO₃ = 1 + 14 + (16 x 3) = 63 g/mol

Mass of HNO₃ = 0.2 x 63 = 12.6 g

Number of moles of H₂O = 2 x 0.05 = 0.1 moles

Molar mass of H₂O = (1 x 2) + 16 = 18 g/mol

Mass of H₂O = 0.1 x 18 = 1.8 g

Mass of the residue = Mass of Pb₃O₄ + Mass of HNO₃ - Mass of H₂O

Mass of the residue = 34.3 g + 12.6 g - 1.8 g

Mass of the residue = 45.1 g

Therefore, the weight of the residue = 45.1 g

User AturSams
by
5.0k points