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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]
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