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)

Be sure to answer all parts. the following equilibrium constants were determined at 1123 k: c(s) + co2(g) ⇌ 2co(g) k'p = 1.30 × 1014 co(g) + cl2(g) ⇌ cocl2(g) k''p = 6.00 × 10−3 calculate the equilibrium constant at 1123 k for the reaction: c(s) + co2(g) + 2cl2(g) ⇌ 2cocl2(g) 4.68 × 10 9 (enter your answer in scientific notation.) write the equilibrium constant expression, kp: