Дано:
Масса оксида фосфора (V) (P2O5) = 35 г
Из уравнения реакции видно, что на один молекулу оксида фосфора (V) требуется 2 атома фосфора и 5 молекул кислорода.
1) Calculating the mass of phosphorus:Molar mass of P2O5 = 2 (atomic mass of P) + 5 (molar mass of O) = 2 31 + 5 16 = 62 + 80 = 142 g/mol
Mass of phosphorus in 35 g of P2O5:(2 atomic mass of P) / molar mass of P2O5 = (2 31) / 142 = 62 / 142 = 0.437 moles
Mass of phosphorus = 0.437 moles (atomic mass of P) = 0.437 31 = 13.5 g
2) Calculating the volume of oxygen at n.t.p.:According to the stoichiometry of the reaction, 1 mol of P2O5 requires 5 mol of O2.
Volume of oxygen gas at n.t.p. in 1 mol:22.4 L/mol
Volume of oxygen gas in 0.437 moles:0.437 mol * 22.4 L/mol = 9.78 L
Therefore, to obtain 35 g of phosphorous oxide (V) (P2O5), we would need 13.5 g of phosphorus and 9.78 L of oxygen gas at n.t.p.
Дано:
Масса оксида фосфора (V) (P2O5) = 35 г
Из уравнения реакции видно, что на один молекулу оксида фосфора (V) требуется 2 атома фосфора и 5 молекул кислорода.
1) Calculating the mass of phosphorus:
Molar mass of P2O5 = 2 (atomic mass of P) + 5 (molar mass of O) = 2 31 + 5 16 = 62 + 80 = 142 g/mol
Mass of phosphorus in 35 g of P2O5:
(2 atomic mass of P) / molar mass of P2O5 = (2 31) / 142 = 62 / 142 = 0.437 moles
Mass of phosphorus = 0.437 moles (atomic mass of P) = 0.437 31 = 13.5 g
2) Calculating the volume of oxygen at n.t.p.:
According to the stoichiometry of the reaction, 1 mol of P2O5 requires 5 mol of O2.
Volume of oxygen gas at n.t.p. in 1 mol:
22.4 L/mol
Volume of oxygen gas in 0.437 moles:
0.437 mol * 22.4 L/mol = 9.78 L
Therefore, to obtain 35 g of phosphorous oxide (V) (P2O5), we would need 13.5 g of phosphorus and 9.78 L of oxygen gas at n.t.p.