Given,
The mass of phosphorus is 8.2 g,
The mass of hydrogen is 4 g,
The molar mass of P₄ is 123.6 g/mol
The molar mass of H₂ is 2.016 g/mol
The molar mass of PH₃ is 33.924 g/mol
The balanced chemical reaction is:
P₄ (s) + 6H₂ (g) → 4PH₃ (g)
First, there is a need to calculate the moles of P₄ and H₂,
Moles of P₄ = mass of P₄/molar mass of P₄ = 8.2 g / 123.6 g/mol
= 0.066 moles
Moles of H₂ = mass of H₂/molar mass of H₂ = 4 g / 2.016 g/mol
= 1.98 moles
From the reaction it is concluded that:
1 mol of P₄ reacts with 6 moles of H₂
0.066 moles of P₄ reacts with 6 × 0.066 = 0.396 moles of H₂.
This signifies that H₂ is in excess amount and P₄ is in limited amount.
Now, there is a need to calculate the moles of PH₃:
As, 1 mole of P₄ reacts to give 4 moles of PH₃
So, 0.066 moles of P₄ reacts to give 4 × 0.066 = 0.264 moles of PH₃.
Now, there is a need to calculate the mass of PH₃
Mass of PH₃ = moles of PH₃ × molar mass of PH₃
Mass of PH₃ = 0.264 moles × 33.924 g/mol = 8.955 g
Therefore, the maximum number of grams of PH₃ formed is 8.955 g.
