Question

proper number of significant figures? 7. reacted with 100.0 mL of 0.333 M HCL The A 1.522 g sample of impure calcium carbonate was reac mpleted required 35.50 ml of 0.300 M NaOH to excess HCL remaining after the reaction was completed regu calcium carbonate? (Molar mass of calcium reach the end point. What is the wt% purity of the calcium carbona carbonate is 100.086 g/mole)

Answer 1

Consider reaction, CaCO ₃ + 2 HCl $\rightarrow$ CaCl ₂ + H₂O + CO ₂

According to reaction, 1 mole calcium carbonate reacts with 2 mole HCl.

Let's calculate moles of HCl consumed by calcium carbonate.

We have, [ HCl ] = No. of moles of HCl / volume of solution in L

No. of moles of HCl = [ HCl ] $\times$ volume of solution in L

$\therefore$ moles of HCl added in the reaction = 0.333 mol / L $\times$ 0.1000 L = 0.0333 mol

moles of NaOH consumed by excess HCl = 0.300 mol / L $\times$ 0.03550 L = 0.01065 mol

$\therefore$ moles of HCl consumed by calcium carbonate = 0.0333 mol - 0.01065 mol = 0.02265 mol

$\therefore$ Moles of calcium carbonate present in the sample = 0.02265 mol HCl $\times$ ( 1 mol CaCO ₃ / 2 mol HCl )

= 0.011325 mol

We have, no. of moles = Mass / Molar mass

$\therefore$ Mass = No. of moles $\times$ Molar mass

Mass of Calcium carbonate in the sample = 0.011325 mol $\times$ 100.086 g /mol = 1.13347 g

% purity of calcium carbonate = ( Actual mass of CaCO ₃ / mass of sample ) $\times$ 100

= ( 1.13347 g / 1.522 g ) $\times$ 100

= 74.47