Answer :
Answer:
mass P4 = 35.998 g
Explanation:
- P4 + 5O2 → P4O10
∴ STP: P = 1 atm; T = 298 K
∴ V O2= 35.5 L
⇒ nO2 = P.V / R.T
∴ R = 0.082 atm.L/K.mol
⇒ nO2 = ((1 atm)×(35.5L))/((0.082 atm.L/K.mol)(298K))
⇒ nO2 = 1.453 mol O2
⇒ mol P4 = (1.453 molO2)×(mol P4/ 5molO2) = 0.2906 mol P4
∴ Mw P4 = 123.895 g/mol
⇒ mass P4 = (0.2906 mol P4)×(123.895 g/mol) = 35.998 g P4
Final answer:
To form solid tetraphosphorus decaoxide, 157.0 grams of phosphorus (P4) react with 35.5L of oxygen (O2) at STP.
Explanation:
To determine the number of grams of phosphorus (P4) that react with 35.5L of oxygen (O2) at STP to form solid tetraphosphorus decaoxide (P4O10), we need to use the stoichiometry of the balanced equation.
The balanced equation is:
4P4 + 5O2 → 2P4O10
From the equation, we can see that 4 moles of phosphorus reacts with 5 moles of oxygen to produce 2 moles of tetraphosphorus decaoxide.
By using the molar volume of a gas at STP (22.4L/mol), we can convert the given volume of oxygen to moles:
35.5L O2 * (1 mol O2 / 22.4L O2) = 1.586 mol O2
Using the stoichiometry of the equation, we can determine the moles of phosphorus:
4 mol P4 * (1.586 mol O2 / 5 mol O2) = 1.268 mol P4
Finally, we can convert the moles of phosphorus to grams:
1.268 mol P4 * (123.89 g P4 / 1 mol P4) = 157.0 g P4
Therefore, 157.0 grams of phosphorus (P4) react with 35.5L of oxygen (O2) at STP to form solid tetraphosphorus decaoxide (P4O10).
Learn more about stoichiometry here:
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