Answer:
B.) P4
Explanation:
Step 1: Convert 25.0 grams of phosphorus into moles of phosphorous:
25 g P4 x 1 mol P4/123.90 g P4 = 0.202 mol P4
Step 2: Convert 0.202 mol P4 into moles of O2 using their stoichiometric ratios:
0.202 mol P4 x 5 mol O2/1 mol P4 = 1.01 mol O2
Step 3: Convert 1.01 mol O2 into grams of O2:
1.01 mol O2 x 31.98 g O2/1 mol O2 = 32.3 g O2
Because 25.0 grams of phosphorous only needs 32.3 grams of oxygen to react, phosphorous will be the limiting reactant, as after the 25.0 grams of phosphorous is used up, there will still be 17.7 grams of oxygen leftover (excess reactant).
Because the reaction cannot occur without more phosphorous to react with the 17.7 grams of oxygen remaining, P4 is the limiting reactant.
Answer: B.) P4
Explanation:
25 g P4 x 1 mol P4/123.90 g P4 = 0.202 mol P4
0.202 mol P4 x 5 mol O2/1 mol P4 = 1.01 mol O2
1.01 mol O2 x 31.98 g O2/1 mol O2 = 32.3 g O2
Because 25.0 grams of phosphorous only needs 32.3 grams of oxygen to react, phosphorous will be the limiting reactant, as after the 25.0 grams of phosphorous is used up, there will still be 17.7 grams of oxygen leftover (excess reactant).
Because the reaction cannot occur without more phosphorous to react with the 17.7 grams of oxygen remaining, P4 is the limiting reactant.
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