Chalk is a mixture of CaCO3 and CaSO4. When added to HCl, only the CaCO3 reacts according to the reaction:
CaCO3(s) + 2 HCl (aq) → CaCl2 (aq) + CO2(g) + H2O (l)
When a 7.54 g piece of chalk is placed in HCl, 2.68 of CO2 is produced. What percentage of the chalk is CaCO3?

The percentage of CaCO₃ in the 7.54 g piece of chalk is 80.8%

We'll begin by calculating the number of mole in 2.68 g of CO₂

Mass of CO₂ = 2.68 g

Molar mass of CO₂ = 12 + (2×16) = 44 g/mol

Mole = mass / molar mass

Mole of CO₂ = 2.68 / 44

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

From the balanced equation above,

1 mole of CaCO₃ reacted to produce 1 mole CO₂.

Therefore,

0.0609 mole of CaCO₃ will also react to produce 0.0609 mole of CO₂

Mole of CaCO₃ = 0.0609 mole

Molar mass of CaCO₃ = 40 + 12 + (16×3) = 100 g/mol

Mass = mole × molar mass

Mass of CaCO₃ = 0.0609 × 100

Mass of CaCO₃ = 6.09 g

Mass of chalk = 7.54 g

Therefore, the percentage of CaCO₃ in the chalk is 80.8%

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Mass percent of CaCO₃ is 81 %

Explanation:

First of all, we determine the moles of CO₂ in order to work with the stoichometry's reaction:

2.68 g . 1mol / 44g = 0.0609 moles

The reaction is: CaCO₃(s) + 2HCl (aq) → CaCl₂ (aq) + CO₂(g) + H₂O (l)

Ratio is 1:1. As 1 mol of CO₂ can be produced by 1 mol of carbonate, the 0.0609 moles of CO₂ must be produced by the same amount.

We convert the moles to mass → 0.0609 mol . 100.08 g /1mol = 6.09 g

So 6.09 g of CaCO₃ are contained in the chalk piece.

The percentage is (mass of CaCO₃ / Mass of chalk) . 100 =

(6.09 g / 7.54 g) . 100 = 81%

This means that (7.54 g - 6.09g ) = 1.45 g of the chalk is CaSO₄

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.