A0.450 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.

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 %

false

because it does not have the potential to be entirely used up

conduction is the transfer of thermal energy (heat in transfer) due to collisions between the molecules in the object. collisions between adjacent atoms and molecules transfer kinetic energy from the warmer to the cooler object. the objects must be in physical contact.