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%

Learn more: link

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

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol