Calcium carbonate is added to separate solutions of hydrochloric acid and ethanoic acid of the same concentration. state one similarity and one difference in the observations you could make.
(i) write an equation for the reaction between hydrochloric acid and calcium carbonate.
(ii) determine the volume of 1.50 mol/dm–3 hydrochloric acid that would react with exactly 1.25 g of calcium carbonate.
(iii) calculate the volume of carbon dioxide, measured at 273 k and 1.01×105 pa, which would be produced when 1.25 g of calcium carbonate reacts completely with the hydrochloric acid.

Explanation:

Similarity - In both the solutions, there will be bubbling and brisk effervescence due to the evolution of carbon dioxide gas.

Difference - Since, HCl is a strong acid than ethanoic acid, it will react more rigorously and rapidly with calcium carbonate.

(i) The equation for reaction between hydrochloric acid and calcium carbonate is shown below:

(ii)

Given, Mass of calcium carbonate = 1.25 g

Molar mass of calcium carbonate = 100.0869 g/mol

The formula for the calculation of moles is shown below:

Thus,

According to the given reaction:

1 mole of calcium carbonate react with 2 moles of HCl.

0.01249 mole of calcium carbonate react with 2 × 0.01249 moles of HCl.

Moles of HCl = 0.02498 moles

Also, considering:

Molarity = 1.50 mol/dm³

So,  

Volume = 0.016653 dm³

1 dm³ = 1 L

Also, 1 L = 1000 mL

So, volume = 16.7 mL

(iii)

1 mole of calcium carbonate forms 1 mole of carbon dioxide

0.01249 mole of calcium carbonate forms 0.01249 mole of carbon dioxide

Given that:

Temperature = 273 K

n = 0.01249 mol

P = 1.01 × 10⁵ Pa

Also, 1 Pa = 1 / 101325  atm

So, P = 1.01 × 10⁵ / 101325  atm = 0.9968 atm

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

0.9968 × V = 0.01249 × 0.0821 L atm/ K mol × 273 K  

⇒V = 0.2808 L