Question

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.

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

How many grams of calcium chloride will be produced when 31.0 g of calcium carbonate is combined with 15.0 g of hydrochloric acid?

How many grams of the excess reactant will remain after the reaction is complete?

Answer 1

1)

Molar mass of CaCO3,

MM = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

mass(CaCO3)= 31.0 g

use:

number of mol of CaCO3,

n = mass of CaCO3/molar mass of CaCO3

=(31 g)/(1.001*10^2 g/mol)

= 0.3097 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 15.0 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(15 g)/(36.46 g/mol)

= 0.4114 mol

Balanced chemical equation is:

CaCO3 + 2 HCl ---> CaCl2 + H2CO3

1 mol of CaCO3 reacts with 2 mol of HCl

for 0.3097 mol of CaCO3, 0.6194 mol of HCl is required

But we have 0.4114 mol of HCl

so, HCl is limiting reagent

we will use HCl in further calculation

Molar mass of CaCl2,

MM = 1*MM(Ca) + 2*MM(Cl)

= 1*40.08 + 2*35.45

= 110.98 g/mol

According to balanced equation

mol of CaCl2 formed = (1/2)* moles of HCl

= (1/2)*0.4114

= 0.2057 mol

use:

mass of CaCl2 = number of mol * molar mass

= 0.2057*1.11*10^2

= 22.83 g

Answer: 22.8 g

2)

According to balanced equation

mol of CaCO3 reacted = (1/2)* moles of HCl

= (1/2)*0.4114

= 0.2057 mol

mol of CaCO3 remaining = mol initially present - mol reacted

mol of CaCO3 remaining = 0.3097 - 0.2057

mol of CaCO3 remaining = 0.104 mol

Molar mass of CaCO3,

MM = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

use:

mass of CaCO3,

m = number of mol * molar mass

= 0.104 mol * 1.001*10^2 g/mol

= 10.41 g

Answer: 10.4 g