What mass of P4 will completely react with 2.50 L of hydrogen gas, at 0 C and 1.50 atm pressure?

Question

Consider the following reaction:

P₄(s) + 6H₂(g) → 4PH₃(g)

What mass of P₄ will completely react with 2.50 L of hydrogen gas, at 0 C and 1.50 atm pressure?

Answer 1

P₄₍s) + 6H₂(g) ==> 4PH₃(g) ... balanced equation

To find the mass of P₄ required, we first need to find moles of H₂ that are present. To do this, we can use the ideal gas law, PV = nRT

P = pressure = 1.50 atm

V = volume = 2.50 L

n = moles = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 0ºC + 273 = 273K

Solving for n (moles) we have...

n = PV/RT = (1.50)(2.50) / (0.0821)(273)

n = 0.167 moles H₂

Now that we have the moles of H2 present, we can use stoichiometry in the balanced equation along with dimensional analysis to find the mass of P4 that will react completely.

0.167 moles H₂ x 1 mol P₄ / 6 moles H₂ x 123.9 g P₄ / mol P₄ = 3.45 g P₄