25 grams of sodium nitrate containing impurities is heated and the oxygen formed during the decomposition

25 grams of sodium nitrate containing impurities is heated and the oxygen formed during the decomposition is used to burn 3.1 grams of phosphorus to phosphorus oxide (V). What is the proportion of pure sodium nitrate in nitrate?

Let’s write down the reaction equations:
2NaNO3 = 2NaNO2 + О2 ↑
5O2 + 4P = P2O5
Let’s start with the second equation:
It can be seen from the reaction equation that:
ν (O2) / 5 = ν (P) / 4
m (P) / 4 * M (P) = V (O2) / 5 * Vm (O2)
Determine the molar mass of phosphorus:
M (P) = 31 g / mol
Determine the volume of oxygen by expressing from the equation:
V (O2) = m (P) * 5 * Vm (O2) / 4 * M (P)
Substitute the numerical values:
V (O2) = m (P) * 5 * Vm (O2) / 4 * M (P) = 3.1 * 5 * 22.4 / 4 * 31 = 2.8 liters.
Now let’s move on to the first reaction:
It can be seen from the reaction equation that:
ν (NaNO3) / 2 = ν (O2)
m (NaNO3) * M (NaNO3) = V (O2) / Vm (O2)
Determine the molar mass of sodium nitrate:
M (NaNO3) = 23 + 14 + 3 * 16 = 85 g / mol
Determine the mass of sodium nitrate by expressing from the equation:
m (NaNO3) = 2 * M (NaNO3) * V (O2) / Vm (O2)
Substitute the numerical values:
m (NaNO3) = 2 * M (NaNO3) * V (O2) / Vm (O2) = 2 * 85 * 2.8 / 22.4 = 21.25 g.
w (NaNO3) = m (NaNO3) * 100% / m (sel) = 21.25 * 100% / 25 = 85%
Answer: the mass fraction of sodium nitrate is 85%.