A 3.31-g sample of lead nitrate, pb (no3) 2, molar mass = 331 g/mol, is heated in an evacuated cylinder with a volume of 2.11 l. the salt decomposes when heated, according to the equation:

Question

Autumn Mclean

Chemistry

29 July, 13:19

A 3.31-g sample of lead nitrate, pb (no3) 2, molar mass = 331 g/mol, is heated in an evacuated cylinder with a volume of 2.11 l. the salt decomposes when heated, according to the equation:

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Answers (1)

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Antwan Goodman · Answer 1

The equation in this problem is: 2Pb (NO3) 2 (s) - - > 2PbO (s) + 4NO2 (g) + O2 (g)

And the question is what the pressure in the cylinder is after decomposition and cooling to a temperature of 300 K.

Solution:

Moles of Pb (NO3) 2 = 3.31/331 = 0.0100

2 moles of Pb (NO3) 2 will decay to mold 4 moles of NO2 and 1 mole of O2. So 0.0100 moles of Pb (NO3) 2 will form 0.02 moles of NO2 and 0.00500 moles of O2

Then use the formula: PV = nRT.

P = (0.02 + 0.005) * 0.082 * 300 / 1.62

= 0.380 atm