This is a limiting reagent problem.
1. Write and balance the equation.
2a. Convert CaCl2 to moles. moles = M x L
2b. Same for K2CO3 (you made a typo in the formula).
3a. Using the coefficients in the balanced equation, convert mol CaCl2 to mol CaCO3.
3b. Same for mol K2CO3 to mol CaCO3.
3c. You arrive at two answers; both can't be correct. The correct answer in limiting reagent problems is ALWAYS the smaller value.
4. Convert the smaller value to grams. g = mol x molar mass.
How many grams of CaCO3 can be precipitated if 50.0 ml of 0.10 M CaCl2 are mixed with 30 ml of 0.2 M KCO3?
1 answer
1 answer