Question

If 4.50 g of HCl are reacted with 15.0 g of CaCO3, according to the following balanced chemical equation, which reactant is going to be the limiting reactant?

(H = 1.01 amu, Cl = 35.45 amu, Ca = 40.08 amu, C = 12.01 amu, O = 16.00 amu)

2 HCl + CaCO3 → CaCl2 + H2O + CO2

Answer 1

Molar mass of HCl = 1*MM(H) + 1*MM(Cl)

= 1*1.01 + 1*35.45

= 36.46 g/mol

mass of HCl = 4.5 g

we have below equation to be used:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(4.5 g)/(36.46 g/mol)

= 0.1234 mol

Molar mass of CaCO3 = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

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

= 100.09 g/mol

mass of CaCO3 = 15.0 g

we have below equation to be used:

number of mol of CaCO3,

n = mass of CaCO3/molar mass of CaCO3

=(15.0 g)/(100.09 g/mol)

= 0.1499 mol

we have the Balanced chemical equation as:

2 HCl + CaCO3 ---> Na2S + 4 CO

2 mol of HCl reacts with 1 mol of CaCO3

for 0.1234 mol of HCl, 6.171*10^-2 mol of CaCO3 is required

But we have 0.1499 mol of CaCO3

so, HCl is limiting reagent

Answer: HCl