What is the maximum number of grams of PH3 that can be formed when 6.2 g of phosphorus reacts with 4.0 g of hydrogen to form PH3?

6.79 g of phosphine can be produced

Explanation:

The reaction is this:

3H₂ + 2P → 2PH₃

We have the mass of the two reactants, so let's find out the limiting reactant, so we can work with the equation. Firstly, we convert the mass to moles (mass / molar mass)

6.2 g / 30.97 g/mol = 0.200 moles of P

4g / 2 g/mol = 2 moles of H₂

Ratio is 3:2.

3 moles of hydrogen react with 2 moles of P

Then, 2 moles of H₂ would react with (2. 2) / 3 = 1.3 moles of P.

We have only 0.2 moles of P, so clearly the phosphorous is the limiting reactant.

Ratio is 2:2. So 2 moles of P can produce 2 moles of phosphine. Therefore, 0.2 moles of P must produce the same amount of phosphine.

Let's convert the moles to mass (mol. molar mass)

0.2 mol. 33.97 g/mol = 6.79 g