Answer:
22.67 L of PH₃
Explanation:
The balanced equation is:
[tex]P_4 (s) + 6H_2(g) \to 4PH_3(g)[/tex]
From the equation:
[tex]34 L \times \dfrac{1 \ mol \ of H_2 }{22.4 \ L \ H_2} \times \dfrac{4 \ mol \ of \ PH_3}{6 \ mol \ H_2} \times \dfrac{22.4 \ L \ PH_3}{1 \ mol \ PH_3}[/tex]
= 22.67 L of PH₃
