Question

1. What volume of F₂ is needed to completely react with excess P4 and produce 1.88 L of PF3?

P4 (s) +6F2 (g) - 4PF3 (g)
0.470 LF2
0.313 LF₂
1.25 LF₂
2.82 LF2

Answer 1

Answer:0.313 L F₂

Explanation:The problem asks for the volume of F₂ needed to react with excess P4 to produce 1.88 L of PF3. To find the answer, we need to balance the chemical equation given: P4 (s) +6F2 (g) - 4PF3 (g).

Since 6 moles of F₂ are required to react with 1 mole of P4 to form 4 moles of PF3, the ratio of F₂ to PF3 is 6:4, or 3:2. So, for every 2 liters of PF3 produced, 3 liters of F₂ are required.

Therefore, if we need 1.88 L of PF3, we must use (3/2)*1.88 L = 2.82 L of F₂.

So, the volume of F₂ needed to completely react with excess P4 and produce 1.88 L of PF3 is 2.82 L.

Answer 2

Answer: 0.313 L F₂

Explanation: The problem asks for the volume of F₂ needed to react with excess P4 to produce 1.88 L of PF3. To find the answer, we need to balance the chemical equation given: P4 (s) +6F2 (g) - 4PF3 (g).

Since 6 moles of F₂ are required to react with 1 mole of P4 to form 4 moles of PF3, the ratio of F₂ to PF3 is 6:4, or 3:2. So, for every 2 liters of PF3 produced, 3 liters of F₂ are required.

Therefore, if we need 1.88 L of PF3, we must use (3/2)*1.88 L = 2.82 L of F₂.

So, the volume of F₂ needed to completely react with excess P4 and produce 1.88 L of PF3 is 2.82 L.