Question

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO, (s)+2 HCI(aq) CaCl, (aq) + H2Oa) + CO, (g) How many grams of calcium chloride will be produced when 32.0 g of calcium carbonate is combined with 14.0 g of hydrochloric acid? mass of CaCl, Which reactant is in excess? O CaCO, O HCI How many grams of the excess reactant will remain after the reaction is complete? mass of excess reactant:

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)= 32.0 g

use:
number of mol of CaCO3,
n = mass of CaCO3/molar mass of CaCO3
=(32 g)/(1.001*10^2 g/mol)
= 0.3197 mol

Molar mass of HCl,
MM = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol


mass(HCl)= 14.0 g

use:
number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(14 g)/(36.46 g/mol)
= 0.384 mol


1 mol of CaCO3 reacts with 2 mol of HCl
for 0.3197 mol of CaCO3, 0.6394 mol of HCl is required
But we have 0.384 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)*0.384
= 0.192 mol


use:
mass of CaCl2 = number of mol * molar mass
= 0.192*1.11*10^2
= 21.31 g
Answer: 21.3 g

2)
HCl is limiting reagent
Answer: HCl

3)
According to balanced equation
mol of CaCO3 reacted = (1/2)* moles of HCl
= (1/2)*0.384
= 0.192 mol
mol of CaCO3 remaining = mol initially present - mol reacted
mol of CaCO3 remaining = 0.3197 - 0.192
mol of CaCO3 remaining = 0.1277 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
= 0.1277 mol * 1.001*10^2 g/mol
= 12.78 g
Answer: 12.8 g