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)
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Expert's answer
2017-03-26T04:59:06-0400
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.
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