# Chemistry help please! and what is STP?

How many grams of phosphine (PH3) can form by the reaction of 33.0 g phosphorus and 90.0 L hydrogen gas at STP?

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

Relevance

STP is standard temperature (273.15K) and pressure of 100 kPa (1 bar).

we need to know which is the limiting reagent and which is in excess. First we need to calculate the number of moles of hydrogen, which can be accomplished by using the ideal gas law.

PV=nRT or n=PV/RT

where P is pressure, Vis volume, R is the Gas Constant and T is temperature

n= 100,000* 0.090/8.314/273.15= 3.963 moles.

We need to calculate the number of moles of phosphorus by dividing its mass by its molar mass= 33/123.9= 0.26634.. so clearly hydrogen is in excess and phosphorus is the limiting reactant.

We know that 1 mole of phosphorus will produce 4 moles of phosphine, and that 0.26634 moles of phosphorus will produce 1.06536 of phosphine. As phosphine has a molar mass of 34, then the mass produced is 34*1.06536= 36.22224 g.