moles O2=31.2/22.4 moles
so you need 1/5 that number moles of P4.
P4(s) + 5 O2(g) -> P4O10(s)
so you need 1/5 that number moles of P4.
1. Calculate the number of moles of O2:
Using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature (STP: 0 degrees Celsius or 273.15 Kelvin).
At STP, the pressure is 1 atm, and the temperature is 273.15 K.
Using the given volume (31.3 L) and these values:
PV = nRT
(1 atm) * (31.3 L) = n * (0.0821 L atm/mol K) * (273.15 K)
n = (1 atm * 31.3 L) / (0.0821 L atm/mol K * 273.15 K)
n ≈ 1.21 mol
2. Use stoichiometry to find the number of moles of phosphorus:
From the balanced equation, we can see that there is a 1:1 mole ratio between P4 and P4O10.
Therefore, 1.21 mol of P4O10 will react with 1.21 mol of P4.
3. Convert moles to grams:
The molar mass of P4 is approximately 123.9 g/mol.
To find the mass of phosphorus, we can use the equation:
mass = moles * molar mass
mass = 1.21 mol * 123.9 g/mol
mass ≈ 149.72 g
Therefore, approximately 149.72 grams of phosphorus will react with 31.3 L of O2 at STP to form tetraphosphorus decaoxide.
Step 1: Convert the volume of O2 from liters to moles.
At STP, 1 mole of any ideal gas occupies 22.4 liters. Therefore, we can calculate the number of moles of O2 as follows:
31.3 L O2 * (1 mole O2 / 22.4 L O2) = 1.398 moles O2
Step 2: Use stoichiometry to determine the mole ratio between O2 and P4.
According to the balanced chemical equation, 1 mole of P4 reacts with 5 moles of O2. Therefore, we can write the following ratio:
5 moles O2 / 1 mole P4
Step 3: Calculate the number of moles of P4.
To determine the moles of P4, we can multiply the moles of O2 by the mole ratio from step 2:
1.398 moles O2 * (1 mole P4 / 5 moles O2) = 0.2796 moles P4
Step 4: Convert moles of P4 to grams.
The molar mass of P4 is 123.88 g/mol. We can calculate the mass of P4 by multiplying the number of moles by the molar mass:
0.2796 moles P4 * (123.88 g P4 / 1 mole P4) ≈ 34.64 g P4
Therefore, approximately 34.64 grams of phosphorus will react with 31.3 L of O2 at STP to form tetraphosphorus decaoxide.