Use the equation below to determine the maximum number of grams of ph3 that can be formed when 8.2 g of phosphorus reacts with 4 g of hydrogen to form ph3

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.

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