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
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