Question

1. If 60.0g of calcium nitrate react with 180.0g of ammonium carbonate, how many grams of calcium carbonate precipitate will be formed?

2. Starting with solid cobalt(II)chloride, how would you prepare 420mL of 0.60M cobalt(II) chloride solution?

Answer 1

D Ca(NO3)2 + (NH4)₂CO3 CaCO3 + 2NH4NO3 moles of Ca(NO3)2 = 60 orglona moles of (NH4)2CO3 = 180 g Anod - 1.873 mol mol moles of calo 3 produced a 0.866 mol mass of CaCO 3 produced a 0.366% 100.08g/mol = 36.63g moles of Collz = molarity xvoll inc) = 0.6Mx 0.42L = 0.252 moles mass of collz required to dissolve in 420 mL of solution. = 0.2524 165.29 g/mol = 41.65 g