Question

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

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

a. How many grams of calcium chloride will be produced when 27.0g of calcium carbonate are combined with 12.0g of hydrochloric acid?

2. Which reactant is excess and how many grams of this reactant will remain after the reaction is complete?

Answer 1

a)

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)= 27.0 g

use:

number of mol of CaCO3,

n = mass of CaCO3/molar mass of CaCO3

=(27 g)/(1.001*10^2 g/mol)

= 0.2698 mol

Molar mass of HCl,

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

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 12.0 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(12 g)/(36.46 g/mol)

= 0.3291 mol

1 mol of CaCO3 reacts with 2 mol of HCl

for 0.2698 mol of CaCO3, 0.5395 mol of HCl is required

But we have 0.3291 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)*0.3291

= 0.1646 mol

use:

mass of CaCl2 = number of mol * molar mass

= 0.1646*1.11*10^2

= 18.26 g

Answer: 18.3 g

2)

HCl is limiting reagent

So, CaCO3 is in excess

According to balanced equation

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

= (1/2)*0.3291

= 0.1646 mol

mol of CaCO3 remaining = mol initially present - mol reacted

mol of CaCO3 remaining = 0.2698 - 0.1646

mol of CaCO3 remaining = 0.1052 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

= 0.1052 mol * 1.001*10^2 g/mol

= 10.53 g

Answer: 10.5 g