On calcining 340 g of sodium nitrate, 33.6 liters of oxygen were obtained. Calculate the mass fraction of impurities

On calcining 340 g of sodium nitrate, 33.6 liters of oxygen were obtained. Calculate the mass fraction of impurities in the nitrate.

1. Calcining sodium nitrate leads to the formation of sodium nitrite and the evolution of oxygen:

2NaNO3 = 2NaNO2 + O2 ↑;

2. Let’s calculate the chemical amount of oxygen:

n (O2) = V (O2): Vm = 33.6: 22.4 = 1.5 mol;

3. Determine the amount of decomposing sodium nitrate:

n (NaNO3) = n (O2) * 2 = 1.5 * 2 = 3 mol;

4. Find the mass of sodium nitrate:

m (NaNO3) = n (NaNO3) * M (NaNO3);

M (NaNO3) = 23 + 14 + 3 * 16 = 85 g / mol;

m (NaNO3) = 3 * 85 = 255 g;

5. Let’s calculate the mass of impurities in nitrate:

m (impurities) = m (nitrate) – m (NaNO3) = 340 – 255 = 85 g;

6. Set the mass fraction of impurities:

w (impurities) = m (impurities): m (nitrate) = 85: 340 = 0.25 or 25%.

Answer: 25%.