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Sagot :
Answer:
60.86 g of PH₃
Explanation:
We'll begin by calculating the number of mole of H₂ that will occupy 60 L. This can be obtained as follow:
22.4 L = 1 mole of H₂
Therefore,
60 L = 60 / 22.4
60 L = 2.68 mole of H₂.
Next, we shall determine the number of mole of PH₃ produced by the reaction of 60 L (i.e 2.68 mole) of H₂. This can be obtained as follow:
P₄ + 6H₂ –> 4PH₃
From the balanced equation above,
6 moles of H₂ reacted to produce 4 moles of PH₃.
Therefore, 2.68 moles of H₂ will react to to produce = (2.68 × 4)/6 = 1.79 moles of PH₃.
Finally, we shall determine the mass of 1.79 moles of PH₃. This can be obtained as follow:
Mole of PH₃ = 1.79 moles
Molar mass of PH₃ = 31 + (3×1)
= 31 + 3 = 34 g/mol
Mass of PH₃ =?
Mass = mole × molar mass
Mass of PH₃ = 1.79 × 34
Mass of PH₃ = 60.86 g
Thus, 60.86 g of PH₃ were obtained from the reaction.
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