Question

Calculate the volume of hydrochloric acid of concentration 1.0 mol / dm3 that is required to react completely with 2.5g of calcium carbonate.

Answer 1

Reaction -

CaCO3 + 2HCl -------> CaCl2 + H2O + CO2 ↑

• From this, it is clear that -

       ➼ 1 part of Calcium Carbonate reacts with 2 parts of HCl to give 1 part of Calcium Chloride, 1 part of water and 1 part of Carbon Dioxide .

Given -

=> 2.5 grams of CaCO3 is to be reacted completely.

=> HCl solution of concentration 1.0 mol/dm3 is used .

Determining the no. of moles of CaCO3 -

(Ca = 40, C = 12, O = 16)

=> The molar mass of CaCO3 = 40 + 12 + 3 * 16 grams per mole .

                                                 => 52 + 48 = 100 grams per mole .

=> Thus, 2.5 grams of CaCO3 is  $\dfrac{2.5}{100}$  moles = 0.025 moles of CaCO3 .

Co-relating this with the equation -

➼ 1 part of CaCO3 = 0.025 moles of CaCO3 .

➼ From the equation, we know that 2 parts of HCl are required to react with 1 part of CaCO3 .

➼ Thus, 2 * 0.025 => 0.05 moles of HCl is required to react completely with 0.025 moles of CaCO3 .

Now -

➼ Concentration of HCl = 1.0 mol per dm3

➼ No. of moles of HCl present = 0.05 moles of HCl

∵ $\boxed{\LARGE{Concentration\:in\:moles\:per\:dm3=\dfrac{No.\: of\: moles\:of\:HCl}{no.\:of\:dm3\:of\:solution} }}$

➼ $\LARGE{1.0 = \dfrac{0.05}{dm3\:of\:solution} }$

➼ dm3 of solution * 1.0 = 0.05

➼ dm3 of solution = 0.05 dm3

∵ 1 dm3 = 1 litre

➼ Volume of solution = 0.05 * 1 litres => 0.05 L

Answer -

=> Thus, the volume of required HCl solution is 0.05 litres .

More to know -

=> Formulae for Molarity, Molality and Normality -

$\LARGE{\boxed{Molarity = \dfrac{No.\:of\:moles\:of\:solute}{no.\:of\:litres\:of\:solution} }\\}$

$\LARGE{\boxed{Molality = \dfrac{No.\:of\:moles\:of\:solute}{Kilograms\:of\:solvent} }}$

$\LARGE{\boxed{Normality =\dfrac{no.\:of\:gram\:eq.}{Litres\:of\:solution} }}$

$\LARGE{\boxed{Normality = Molarity * N-Factor}}$

=> Here -

$\LARGE{No.\:of\:gram\:equivalents = \dfrac{Mass\:of\:solute}{Eq.\:mass\:of\:solute} }$

$\LARGE{Equivalent\:mass\:of\:solute = \dfrac{Molar\:mass\:of\:solute}{N-Factor}}$

Answer 2

Answer is in the attachment