28.4 grams of P₂O₅ will be obtained from the interaction of 11.2 liters of oxygen with phosphorus.
Let's consider the following balanced equation.
4 P + 5 O₂ ⇒ 2 P₂O₅
Since the conditions are not specified, we will assume that we are working at standard temperature and pressure. At STP, 1 mole of an ideal gas occupies 22.4 L. The volume of 11.2 L of oxygen at STP, assuming ideal behavior, is:
[tex]11.2 L \times \frac{1mol}{22.4L} = 0.500 mol[/tex]
The molar ratio of O₂ to P₂O₅ is 5:2. The moles of P₂O₅ obtained from 0.500 moles of O₂ are:
[tex]0.500 mol O_2 \times \frac{2molP_2O_5}{5molO_2} =0.200 mol P_2O_5[/tex]
The molar mass of P₂O₅ is 141.94 g/mol. The mass corresponding to 0.200 moles of P₂O₅ is:
[tex]0.200 mol \times \frac{141.94 g}{mol} = 28.4 g[/tex]
28.4 grams of P₂O₅ will be obtained from the interaction of 11.2 liters of oxygen with phosphorus.
You can learn more about stoichiometry here: https://brainly.com/question/22288091
