0.479g of impure CaCO 3 is dissolved im 50.0mL of0.131mol/L HCl(aq), which is in

Question

0.479g of impure CaCO3 is dissolved im 50.0mL of0.131mol/L HCl(aq), which is in excess. The excess HCl(aq) istitrated with 6.80mL of 0.110mol/L NaOH(aq). Assuming that thesample contains noting other than CaCO3 that reactscalculate the percentage CaCO3 in the sample. 0.479g of impure CaCO3 is dissolved im 50.0mL of0.131mol/L HCl(aq), which is in excess. The excess HCl(aq) istitrated with 6.80mL of 0.110mol/L NaOH(aq). Assuming that thesample contains noting other than CaCO3 that reactscalculate the percentage CaCO3 in the sample.

Explanation / Answer

CaCO3 + 2HCl -> CaCl2 +H2CO3 HCl + NaOH -> NaCl + H2O c(NaOH) = n/v n(NaOH) = cV = 0.11 x 6.8 x 10-3 = 7.48 x10-4 mol n(HCl) in excess = 7.48 x 10-4 mol c(HCl) before reacting with NaOH = n/v n(HCl) before reacting with NaOH = cv = 0.131 x 0.05 = 6.55 x10-3 mol n(HCl) reacted with CaCO3 = 6.55 x 10-3 -7.48 x 10-4 = 5.802 x 10-3 mol n(CaCO3) = 1/2 x n(HCl) = 2.901 x 10-3mol n(CaCO3) = m/M m(CaCO3) = nM = 2.901 x 10-3 x 100 =0.2901g %CaCO3 in the sample = 0.2901/0.479 x 100 =60.56% Hope this helps!

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