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

Answer 1

Okay first lets figure out the equation
P+H2---->PH3
balance it now
we have: 2P + 3H2 --->2PH3
equation is now balanced.

let us now find the reactant in excess
2 moles of P= 3 moles of H2
2(31)g of P = 3(2)g of H2
6.2 g of P= x

x= 6.2x6/62= 0.6 g of H2

4 grams of H2 are reacting which proves that H2 is the reactant in excess since only 0.6 grams were needed.

So we use the limited reactant P to find out the mass of PH3 formed.
6.2 g of P = x
2(31)g of P= 2(31+3g) of PH3

68 g x 6.2g/62 g
= 6.8 g of PH3
I hope I am correct

Answer 2

The answer by mohamed is correct except your teacher may have instructed you to write P as P4. If so that will change the equation but it will not change the number of grams PH3 formed.