Question

< Question 1 of 3 > 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) + 2 HCl(aq) → CaCl, (aq) + H2O(1) + CO2(g) The excess HCl(aq) is titrated by 8.45 mL of 0.125 M NaOH(aq). Calculate the mass percentage of CaCO,(s) in the sample. mass percentage:

Answer 1

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S0) now hoe to Caltulat herema of o. 1som) Jata Voluneof Hl-(So-9funnew Strengje of. Hee a. 150 M vojume of Naot requhed 8.45 m Strang of Nao of trom due Aaw of acidimehy alealimty O.125 (m) Nos4 VILel Vwae rSaeu el 8.4 ml xo. 12s O.ISO ) .04 mL Coted veh 25g no) AS We assumed that he emin vomee Hel is o-) arbitgo So- 1.04 mL SO-1.04= 42.96 mL The 42.26mmL of o.150n) Hel is Consed byeamma 0f Ca Cos Present n the sonmele As we Knoos Ca Co2 Lo0om 1m) HUsaSom of -0. 3222 9m, CS canaedtwth Ca cos presend mthe Samelt fs a.3222 CamScanner

t0tak Amountof Samele toaken o. 4so . a3 222 m 0.4907m wm) * Calos mars parcentaje of 02) 71.69%0 (09 (s)2810 2x b2x 14 bZ 02S1.0JA 2 .3 Scanned with CS CamScanner
the mas percentage of ccalcium carbonate in the given sample is 71.6%.