Step 1
The equation provided here:
8 P4 + 3 S8 ⇨ 8 P4S3
--------------
Step 2
Information provided:
75.0 g of P4
55.0 g of S8
-----
Information needed:
The molar masses of:
P4) 123.8 g/mol
S8) 256.5 g/mol
(please, remember that the periodic table is useful here)
--------------
Step 3
Procedure:
By stoichiometry,
8 P4 + 3 S8 ⇨ 8 P4S3
8 x 123.8 g P4 ------------ 3 x 256.5 g S8
75.0 g P4 ------------- X
X = 75.0 g P4 x 3 x 256.5 g S8/8 x 123.8 g P4
X = 52.3 g S8
For 75.0 g of P4, 52.3 g of S8 is needed, but there is 55.0 g of S8, Therefore, the excess reactant is S8 and the limiting reactant P4.
Answer:
The limiting reactant = P4
The excess reactant = S8
