A student adds 50.0 mL of 0.25 M K2SO4 solution to 25.0 mL 0.50 M BaCl2 solution. What is the
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Chemistry, 06.05.2020 01:07 anonwarrior
A student adds 50.0 mL of 0.25 M K2SO4 solution to 25.0 mL 0.50 M BaCl2 solution. What is the
mass of the precipitate that can be recovered if the percent yield of the reaction is 75%? Ba = 137;
Cl = 35.5; S = 32 O = 16; K = 39. Note: You must write the balanced equation first.
BaCl2(aq) + K2SO4(aq)  BaSO4(s) + 2KCl(aq)
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