= 0.06390 mol Ba
= 0.1130 mol S
When barium (Ba) reacts with sulfur (S) to form barium sulfide (BaS), each Ba atom reacts with an S atom. If 2.50 cm3 of Ba reacts with 1.75 cm3 of S, are there enough Ba atoms to react with the S atoms (d of Ba = 3.51 g/cm3; d of S = 2.07 g/cm3)?
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