Calculate the total volume of gases (l) obtained by heating lead nitrate (2), if the mass is 202.4 g of a solid product.

Let’s write down given:

Pb (NO3) 2 – heated

m (solid) = 202.4 g

v (gases) -?

Decision.

1) Let us write the reaction equation for the decomposition of lead (II) nitrate:

Pb (NO3) 2 = Pb O + NO2 + O2

Let’s equalize the reaction:

2 Pb (NO3) 2 = 2 Pb O + 4 NO2 + O2

The solid residue, known from the problem statement, is lead (II) oxide – PbO, gaseous products – NO2 (nitrogen oxide (IV)) and oxygen – O2.

2) Calculate the mass of lead (II) nitrate that has entered into the reaction:

M (Pb (NO3) 2) = 207 * 1 + (14 * 1 + 16 * 3) * 2 = 331 g / mol,

M (Pb O) = 223 g / mol

From the reaction equation we see that

2 * 331 g (Pb (NO3) 2) forms 2 * 223 g (Pb O)

x g (Pb (NO3) 2) forms 202.4 g (Pb O)

x = 202.4 * 2 * 331/2 * 223

x = 300.4 g (Pb (NO3) 2)

3) Calculate the volume of nitrogen oxide released as a result of the reaction:

Vm = 22.4 l / mol

2 * 331 g (Pb (NO3) 2) – 4 * 22.4 l (NO2)

300.4 g (Pb (NO3) 2) – x L (NO2), hence

x l (NO2) = 300.4 g * 4 * 22.4 l / 2 * 331 g

x (NO2) = 40.66 l (NO2)

4) Calculate the volume of oxygen released as a result of the reaction:

2 * 331 g (Pb (NO3) 2) – 22.4 L (O2)

300.4 g (Pb (NO3) 2) – u l (O2), hence

Y l (O2) = 2 * 331 g * 22.4 l / 2 * 331 g

Y = 10.2 l (O2)

5) Calculate the total volume of gases released as a result of the reaction:

v (NO2) + v (О2) = 40.66 l + 10.2 l = 50.86 l

Answer: the volume of gases was 50.86 liters.