5. A 5.63 g sample of calcium ore was dissolved in HCl and gravimetrically analyzed, through the precipitation of calcium into CaC2O4 · H2O. The precipitate was filtered, washed, dried and ignited at 500 deg C until the weight was constant, giving a final mass of 2.35 grams pure CaCO3 (100.087 g/mol).
Calculate the % Calcium (40.078 g/mol) in the sample.
6. For the reaction HCN (aq) + H2O (l) CN- (aq) + H3O+ (aq), all of the following will shift the equilibrium toward the products, except:
I. adding HCN
II. adding H2O
III. removing CN−
IV. removing H3O+
5. Theoritically : 1 mole CaCO3 is giving 1 mole of Ca:
n(Ca) = n(CaCO3) = m/M = 2.35g/100.087g/mol = 0.02348 mol
m(Ca) = 0.02348mol*40.078g/mol = 0.941 g
% Calcium in the sample = 0.941*100%/5.63 = 16.7%
6. II. adding H2O