What is the volume of nitrogen dioxide (in l) is produce from the complete reaction of 16.87g of lead (ii) nitrate at stp?

Molar mass:

NO2 = 46.0 g/mol
Pb(NO3)2 = 331.2 g/mol

first calculates the number of moles of NO2 through the stoichiometry of reaction:

2 Pb(NO3)2(g) > 2 PbO(s) + 4 NO2(g) + O2(g)

2 x 331.2 g > 4 x 46.0 g
16.87 g > ? ( mass of NO2)

mass of NO2 = 16.87 x 4 x 46.0 / 2 x 331.2

mass of NO2 = 3104.08 / 662.4

= 4.686 g of NO2

Number of moles:

1 mole NO2 46.0 g
? moles NO2 4.686 g

4.686 x 1 / 46.0 => 0.101 moles of NO2

1 mole 22.4 L ( at STP)
0.101 moles ?

0.101 x 22.4 / 1

=>  2.26 L

hope this helps!

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