Calculate the amount of all substances participating in the reaction 4Ρ + 5Ο₂ = 2Ρ₂Ο₅

Calculate the amount of all substances participating in the reaction 4Ρ + 5Ο₂ = 2Ρ₂Ο₅, if 310 g of phosphorus has reacted.

Let’s implement the solution:
1. Based on the data, we write down the process:
m = 310 g. Y -? Y -?
4P + 5O2 = 2P2O5 – compounds, phosphorus oxide was obtained (5);
2. Set the values:
M (P) = 30.9 g / mol;
M (O2) = 32 g / mol;
M (P2O5) = 141.8 g / mol.
3. Define Y of the starting material:
Y (P) = m / M = 310 / 30.9 = 10 mol.
4. Proportions:
10 mol (P) – X mol (O2);
-4 mol -5 mol from here, X mol (O2) = 10 * 5/4 = 12.5 mol (substance in excess).
Calculations are made taking into account the deficient substance.
10 mol (P) – X mol (P2O5);
-4 mol -2 mol from here, X mol (P2O5) = 10 * 2/4 = 5 mol.
Answer: phosphorus oxide (5) was obtained in the amount of 5 mol, oxygen in the amount of 12.5 mol is required.