How many grams of iron III sulfate form when 62.0 mL of 0.160 M of Barium chloride reacts with 58.0 ml of 0.065 M os Sodium Sulfate? Please show the steps necessary to reach the answer as well. This question is not in the textbook solutions section of chegg.
moles of BaCl2 = 62 x 0.160 / 1000 = 9.92 x 10^-3
moles of Na2SO4 = 58 x 0.065 / 1000 = 3.77 x 10^-3
BaCl2 (aq) + Na2SO4 (aq) -------------> BaSO4 (s) + 2 NaCl (aq)
1 1
9.92 x 10^-3 3.77 x 10^-3
here limiting reagent Na2SO4.
1 mol Na2SO4 --------------> 1 mol BaSO4
3.77 x 10^-3 mol Na2SO4 -----------> ??
moles of BaSO4 formed = 3.77 x 10^-3
mass of BaSO4 formed = 233.39 x 3.77 x 10^-3
= 0.880 g
Get Answers For Free
Most questions answered within 1 hours.