How many grams of product, PH3(gas) can form when 37.5g of phosphorus in the form of P4(s) and 83.0L of H2(g) react at STP?

How many grams of product, PH3(gas) can form when 37.5g of phosphorus in the form of P4(s) and 83.0L of H2(g) react at STP? ( Hint limiting reactants and P4 represents a molecular form of Payment) and STP= 1 atm and 273K.

P4(s) + 6 H2(g)—-> 4 PH3 (g)

ThusSince , PisThus,But we have,Molar mass of P.P4 CS)In the balanced equation,Hence, 37 50 of P4 – 37 59of t . Thw , P441’07+ 6 12 Ci -at ST.P , 1 mot #2amount of PH, producedP4 ‘….