How many grams of solid barium sulfate form when?

Richard C - 2008-09-23 11:49:04 - Chemistry

How many grams of solid barium sulfate form when 16.9 mL of 0.160 M barium chloride reacts with 65.5 mL of 0.055 M sodium sulfate? Aqueous sodium chloride is the other product. thank you very much Dr. A

Best Answer:

Moles BaCl2 = 0.0169 L x 0.160 M = 0.00270 Moles Na2SO4 = 0.0655 L x 0.055 M = 0.00360 BaCl2 + Na2SO4 >> BaSO4 (s) + 2 NaCl the ratio is 1 : 1 so BaCl2 is the limiting reactant we would get 0.00270 moles of BaSO4 mass BaSO4 = 0.00270 mol x 233.43 g/mol =0.630 g

Answers:

Moles BaCl2 = 0.0169 L x 0.160 M = 0.00270 Moles Na2SO4 = 0.0655 L x 0.055 M = 0.00360 BaCl2 + Na2SO4 >> BaSO4 (s) + 2 NaCl the ratio is 1 : 1 so BaCl2 is the limiting reactant we would get 0.00270 moles of BaSO4 mass BaSO4 = 0.00270 mol x 233.43 g/mol =0.630 g