Question

Limestone (CaCO3, 100.09 g/mol) is present in many sedimentary rocks. Suppose the limestone in a 5.000 g rock sample is dissolved in hydrochloric acid, and then 33.30 mL of 0.1500 M Na2C2O4 is added to precipitate the Ca2+ ions as CaC204 (128.10 g/mol). What is the mass percent limestone in the original rock sample? 10.00% 12.80% 17.35% 85.31% 66.67%

Answer 1

given data are Cacoz = 100.09 g/mol mass of socks soooog. 33.30m L of 0.1500 m Nazcz0y = 33-30m LX 10-3L x 0.1500 mol imL . = 4.995 X10-3 mol ca2t+ Noz coz → Cacoz 50 Imol of Naz Loz gives i mol of CaCO3 or 4.995 X183 mal will give = Imalcaco3 x 4.9958 Calo3 x 4.995710 ml NaCO3 mol Na Coz = 4.995X10-3 mol Calozi now 4.995X10 Imol Cacz = 4.995 X10-3 mol X100.098/ml = 0.59 % slimestone = mass of limestone , mestone x 100 = Oost 4100 = 10.00% enrock Total mass Sobogtros = 10.00%