How many grams of phosphorus (P4) are needed to completely consume 80.8 L of chlorine gas according to the following reaction at 25 °C and 1 atm?
phosphorus (P4) ( s ) + chlorine ( g ) phosphorus trichloride ( l )
The balanced chemical equation for the reaction is:
P4 (s) + 6 Cl2 (g) -> 4 PCl3 (l)
According to the stoichiometry of the reaction, 1 mole of P4 reacts with 6 moles of Cl2 to produce 4 moles of PCl3.
Now we need to convert the volume of Cl2 gas to moles using the ideal gas law:
PV = nRT
n = PV/RT
n = (1 atm) * (80.8 L) / (0.0821 L.atm/mol.K * 298 K)
n = 3.4 moles
Since 1 mol of P4 reacts with 6 moles of Cl2, we would need (3.4 moles Cl2 / 6) = 0.567 moles of P4 to completely consume 80.8 L of Cl2 gas.
The molar mass of P4 is approximately 123.9 g/mol.
Therefore, the mass of P4 needed would be:
0.567 moles * 123.9 g/mol = 70.33 grams of P4.
Hence, 70.33 grams of P4 are needed to completely consume 80.8 L of Cl2 gas.