For the reaction CaCO3(s)+ 2HCl(aq)= CaCl2(aq)+ CO2(g)+ H2O(l) how many grams of CaCl2 can be obtained if 14.6 g HCl is allowed in CaCO3?
To calculate the grams of CaCl2 that can be obtained in the reaction, we need to use the concept of stoichiometry. Stoichiometry allows us to determine the quantitative relationships between the reactants and products in a chemical reaction.
First, we need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed. In this case, we are given the mass of HCl (14.6 g) and the formula mass of CaCO3, HCl, and CaCl2.
1. Calculate the formula mass of HCl:
HCl = 1(atomic mass of H) + 1(atomic mass of Cl)
= 1(1.007 g/mol) + 1(35.453 g/mol)
= 36.460 g/mol
2. Calculate the number of moles of HCl:
moles of HCl = mass of HCl / formula mass of HCl
moles of HCl = 14.6 g / 36.460 g/mol
= 0.4004 mol
3. Next, we need to determine the stoichiometric ratio of HCl to CaCl2. Looking at the balanced chemical equation:
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
The stoichiometric ratio between HCl and CaCl2 is 2:1. This means that for every 2 moles of HCl, we obtain 1 mole of CaCl2.
4. Determine the moles of CaCl2 that can be obtained:
moles of CaCl2 = moles of HCl / (stoichiometric ratio of HCl to CaCl2)
moles of CaCl2 = 0.4004 mol / (2 moles of HCl / 1 mole of CaCl2)
= 0.2002 mol
5. Finally, calculate the mass of CaCl2:
mass of CaCl2 = moles of CaCl2 × formula mass of CaCl2
mass of CaCl2 = 0.2002 mol × (calculation of formula mass of CaCl2)
The formula mass of CaCl2 is:
CaCl2 = (atomic mass of Ca) + 2(atomic mass of Cl)
= (40.08 g/mol) + 2(35.453 g/mol)
= 110.98 g/mol
mass of CaCl2 = 0.2002 mol × 110.98 g/mol
= 22.203 grams
Therefore, approximately 22.203 grams of CaCl2 can be obtained if 14.6 grams of HCl is allowed to react with CaCO3.
To calculate the grams of CaCl2 that can be obtained, we need to determine the limiting reactant. The limiting reactant is the reactant that gets completely consumed and determines the maximum amount of product that can be formed.
First, we need to convert the given mass of HCl to moles:
Given: Mass of HCl = 14.6 g
To convert grams of HCl to moles, we need to use its molar mass. The molar mass of HCl is:
Molar mass of HCl = 1.007 g/mol (hydrogen) + 35.453 g/mol (chlorine) = 36.460 g/mol
Moles of HCl = Mass of HCl / Molar mass of HCl
Moles of HCl = 14.6 g / 36.460 g/mol
Moles of HCl ≈ 0.400 mol
According to the balanced chemical equation, the stoichiometric ratio between HCl and CaCl2 is 2:1. This means that for every 2 moles of HCl, we get 1 mole of CaCl2.
From the stoichiometry, we can determine the moles of CaCl2 that can be obtained:
Moles of CaCl2 = Moles of HCl × (1 mole CaCl2 / 2 moles HCl)
Moles of CaCl2 = 0.400 mol × (1 mol CaCl2 / 2 mol HCl)
Moles of CaCl2 = 0.200 mol
Finally, we need to convert the moles of CaCl2 to grams. The molar mass of CaCl2 is:
Molar mass of CaCl2 = 40.078 g/mol (calcium) + 35.453 g/mol (chlorine) + 35.453 g/mol (chlorine) = 110.984 g/mol
Mass of CaCl2 = Moles of CaCl2 × Molar mass of CaCl2
Mass of CaCl2 = 0.200 mol × 110.984 g/mol
Mass of CaCl2 ≈ 22.197 g
Therefore, approximately 22.197 grams of CaCl2 can be obtained if 14.6 grams of HCl is allowed to react with CaCO3.
Do you mean is 14.6 g CaCO3 is formed?
mols HCl = grams/molar mass
Convert mols HCl to mols CaCl2.
Convert mols CaCl2 to grams. g = mols x molar mass.