PF₃ reacts with XeF₄ to give PF₅. In theory, how many moles of PF₅ can be produced from 100.0 g of PF₃ and 50.0 g of XeF₄?
2 PF₃(g) + XeF₄(s) $\rightarrow$ 2 PF₅(g) + Xe(g)

Question

PF₃ reacts with XeF₄ to give PF₅. In theory, how many moles of PF₅ can be produced from 100.0 g of PF₃ and 50.0 g of XeF₄? 2 PF₃(g) + XeF₄(s) $\rightarrow$ 2 PF₅(g) + Xe(g)

Quizplus · Accepted Answer

The Answer of PF₃ reacts with XeF₄ to give PF₅. In theory, how many moles of PF₅ can be produced from 100.0 g of PF₃ and 50.0 g of XeF₄? 2 PF₃(g) + XeF₄(s) $\rightarrow$ 2 PF₅(g) + Xe(g)

PF3 Reacts with XeF4 to Give PF5 →\rightarrow→ 2 PF5(g) + Xe(g) A) 0

PF₃ Reacts with XeF₄ to Give PF₅ $\rightarrow$ 2 PF₅(g) + Xe(g)
A) 0