Question

Dissolving 3.05 g of an impure sample of CaCO₃ in an excess of HCl acid produced 0.636 L of CO₂ (measured at 25°C and 792 mmHg).

Calculate the percent by mass of CaCO₃ in the sample.
State any assumptions.

Answer 1

88.8 % by mass of CaCO₃

Step-by-step explanation:

This is the reaction:

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Let's determine the moles of formed gas, with the Ideal Gases Law

P . V = n . R .T

Firsty we do this convesions:

- T → 25°C + 273 = 298K

- P → It is in mmHg, it must be in atm → 792 mmHg . 1 atm / 760 mmHg = 1.04 atm

1.04 atm . 0.636L = n . 0.082 . 298K

(1.04 atm . 0.636L) / (0.082 . 298K) = 0.0271 moles

Ratio is 1:1, so 0.0271 moles of dioxide were produced by 0.0271 moles of carbonate. Let's convert the moles to mass

0.0271 mol . 100.08 g / 1mol = 2.71 g

This is the mass of salt, that has reacted. Let's calculate the percent by mass.

(Mass of salt / Total mass) . 100 =

(2.71 g / 3.05 g) . 100 = 88.8 %