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 ) + 2 HCl ( aq ) ⟶ CaCl2 ( aq ) + H2O ( l ) + CO2 ( g ) The excess HCl ( aq ) is titrated by 5.35 mL of 0.125 M NaOH ( aq ) . Calculate the mass percentage of CaCO3(s) in the sample.

Answer 1

Answer:

The mass percentage of CaCO3(s) in the sample is 83.4%

Explanation:

CaCO₃ + 2HCl  →  CaCl₂  +  H2O  +  CO₂

Let's find out the moles of each reactant:

[HCl] = 0.150 mol/L

Molarity . volume = moles

0.150 m/L . 0.050L = 7.5x10⁻³ moles HCl

Molar mass of CaCO₃ =100.08 g

Moles of CaCO₃ = Mass / Molar mass

0.450 g/100.08 g = 4.49x10⁻³ moles of salt

Ratio is 1:2

If 2 moles of HCl react with 1 mol of salt

7.5x10⁻³ moles of HCl react with 7.5x10⁻³ / 2 = 3.75x10⁻³ moles

We must find out the mass

Moles of CaCO₃ . Molar mass CO₃ = 0.3753 g

So now, the mass percentage of salt will be

0.450g ___ 100%

0.3753 g ____ (0.3753 .100 )/0.450 =83.4 %