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)= 28.0 g
use:
number of mol of CaCO3,
n = mass of CaCO3/molar mass of CaCO3
=(28 g)/(1.001*10^2 g/mol)
= 0.2797 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
1 mol of CaCO3 reacts with 2 mol of HCl
for 0.2797 mol of CaCO3, 0.5595 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)
HCl is limiting reagent
Answer: HCl
3)
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 = 0.2797 - 0.2057
mol of CaCO3 remaining = 7.403*10^-2 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
= 7.403*10^-2 mol * 1.001*10^2 g/mol
= 7.41 g
Answer: 7.41 g
< Question 2 of 17 > Attempt 2 - When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon d...
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When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. Caco, (s) + 2 HCl(aq) +CaCl(aq) + H2O(1) + CO,() How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate is combined with 10.0 g of hydrochloric acid? mass of CaCl,:|| TOOLS Which reactant i x10' OHCI Caco How many grams of the excess reactant will remain after the reaction is complete? mass of excess reactant:
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