Question

A 0.450 g sample of impure CaCO3(s) is dissolved in 50.0 mL of 0.150 M HCl(aq) . The equation for the reaction is CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+H2O(l)+CO2(g) The excess HCl(aq) is titrated by 9.05 mL of 0.125 M NaOH(aq) . Calculate the mass percentage of CaCO3(s) in the sample.

Answer 1

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Calo (s) + 2 HCl cag - Call caq) + H2O(gs+ CO₂(g) → 0 Hellcaq) + NaOH (69) Nach (aq) + H₂0 (1) Excers stad till es têbrated with Madhe. NaOH: Molarity = 0.125M = 0.125 molt ; volume = 9.05mL = 0.009052 Molarity = no. of moles (mol) n=MXV = (0.125 mol/L) (0.00905L) I volume of solution (L) n= 0.00113 mot From equation , 1 mol Hel reacts with 1 mol NowH (lll med ratio) Su 0.00113 mol Hel reacts with 0.00113 mol NaOH. and calculate the no. of moles of Hel initially added to caco. n=Mxr = (0.150 mol/L) (50.0mL) = 6.150 mol/L) Co.0506) - 0.0075 mol out of 0.0075 mol of Hello 0.00113 mol is excess (not reacted with Caco 3) So All Creacted) = Hal Ecotal - Hll cercess) - 0.0075 mol-0.00113 mol 0.00637m of Now, from equation , 1 mol cacos reacts with 2 mol HCl (1:e mat ratio) so – mol cacaz reacts with 0.00637 mot tee. amol = 0.003185mos of caco, Mass of cacoa = no. of moles x molar mass = 0.003185 max 100.0 g/mol = 0.31859 of caco, # Mass %. = mass of Caco, 0.31858 mass of sample » Imolx 0.00637 mol ( mol mass molar masy - 100= 0.4 50 x 100 = 70.8%