Question

Calcium carbonate reacts with aqueous \[\ce{HCl}\] to give \[\ce{CaCl2}\] and \[\ce{CO2}\] according to the reaction given below:

\[\ce{CaCO3(s) + 2HCl (aq) -> CaCl2 (aq) + CO2(g) + H2O (l)}\]

What mass of \[\ce{CaCl2}\] will be formed when 250 mL of 0.76 M \[\ce{HCl}\] reacts with 1000 g of \[\ce{CaCO3}\]? Name the limiting reagent. Calculate the number of moles of \[\ce{CaCl2}\] formed in the reaction.

Answer 1

The volume of \[\ce{HCl}\] solution is 250 mL and its molarity is 0.76 M.

Calculate the number of moles of \[\ce{HCl}\] as follows,

Moles of \[\ce{HCl}\] = Molarity × Volume (in L)

= 0.76 M × 0.250 L

= 0.19 mol

The molar mass of \[\ce{CaCO3}\] is 100 g/gQl and the mass of \[\ce{CaCO3}\] is igven as 1000 g.

The number of moles of \[\ce{CaCO3}\] is calculated as

Moles of \[\ce{CaCO3}\] = Mass/Molar mass

= `(1000  g)/((100  g)/(mol)`

= 10 mol

To the given reaction 1 mole of \[\ce{CaCO3}\] requires 2 moles of \[\ce{HCl}\]. So, the required number of moles of \[\ce{HCl}\] for 10 moles of \[\ce{CaCO3}\] is calculated as

Moles of HCl = 2 mol of  \[\ce{HCl}\]/1 mol of \[\ce{CaCO2}\] × 10 mol of \[\ce{CaCO3}\]

= 20 mol

So, the required number of moles of \[\ce{HCl}\] is 20 mol but only 0.19 mol are given. So, \[\ce{HCl}\] is a limiting reagent. So, the amount of calcium chloride formed is depend upon the limiting reagent, that is , the amount of \[\ce{HCl}\] available.

To the reaction, 2 moles of \[\ce{HCl}\] gives 1 pol of \[\ce{CaCl2 . Sg}\], the number of moles of calcium chloride produced by 0.19 pol of \[\ce{HCl}\] as follows,

Moles of \[\ce{CaCl2}\] = 1 mol of \[\ce{CaCl2}\]/2 mol of HCl × 0.19 mol of HCl

= 0.095 mol

The molar mass of calcium chloride is 111 g/mol.

So, itis mass is calculated as,

Mas of \[\ce{CaCl2}\] = Moles  Molar mass

= 00.095 mol × 11 g/mol

= 10.54 g