Answer to Question #66733 in General Chemistry for Lynda

Question

How many grams of Barium metal are needed to produce 75 grams of barium sulfide in the following equation: Ba(s) + S8 (s) + BaS (s)

Answer 1

The chemical equation is
8Ba(s) + S8 (s) → 8BaS (s)

Molar mass of Barium is M = 137.3 g/mol
Molar mass of barium sulfide is M = 169.4 g/mol
Using the chemical reaction between Ba and S8:
X g ----- 75 g
8Ba(s) + S8 (s) → 8BaS (s)
8∙137.3 ----- 8∙169.4
We can determine the mass of barium, according to proportion:
X/(8∙137.3) = 75/(8∙169.4)
Where X is mass of Ba:
X = (75∙8∙137.3)/(8∙169.4) = 60.8 (g)
So, the mass of barim equals 60.8 grams.
Answer: m(Ba) = 60.8 g.