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Use the following balanced reaction to solve:
P4 (s) + 6H2 (g) → 4PH3 (g)

How many grams of phosphorus trihydride will be formed by reacting 60 L of Hydrogen gas with an excess of P4?


Sagot :

Answer:  60.7 g of [tex]PH_3[/tex] will be formed.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}[/tex]    

[tex]\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles[/tex]

The balanced chemical reaction is

[tex]P_4(s)+6H_2(g)\rightarrow 4PH_3(g)[/tex]

[tex]H_2[/tex] is the limiting reagent as it limits the formation of product and [tex]P_4[/tex] is the excess reagent.

According to stoichiometry :

6 moles of [tex]H_2[/tex] produce = 4 moles of [tex]PH_3[/tex]

Thus 2.68 moles of [tex]H_2[/tex] will produce=[tex]\frac{4}{6}\times 2.68=1.79moles[/tex]  of [tex]PH_3[/tex]

Mass of [tex]PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g[/tex]

Thus 60.7 g of [tex]PH_3[/tex] will be formed by reactiong 60 L of hydrogen gas with an excess of [tex]P_4[/tex]