Question

What is the volume of nitrogen dioxide (in L) is produce from the complete reaction of 16.87g of lead (II) nitrate at STP?

Answer 1

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!