Formula: CaCO3(s) + HCl(aq) --> CO2(g) + CaCl2(aq) + H2O
How many grams of CO2 can be made by reacting 125 g of CaCO3 with 125 grams of HCl?
CaCO3(s) + 2HCl(aq) --> CO2(g) + CaCl2(aq) + H2O
Note that I balanced the equation.
This is a limiting reagent (LR) problem. You know that when amounts are given for BOTH reactants instead of one reactant and an excess of the other one. So this essentially is two stoichiometry problems in one. I do these the long way but there is a shorter way.
mols CaCO3 = g/molar mass = 125/100 = 1.25
mols HCl = 125/36.5 = 3.42
Now convert each of these into how many mols CO2 would be formed IF the other reagent was in excess.
For CaCO3 that will be 1.25 x (1 mol CO2/1 mol CaCO3) = 1.25 mols CO2.
For HCl that will be 3.42 x (2 mols HCl/1 mol CO2) = 3.42 x 2 = 6.84 mols CO2.
In LR problems the SMALLER number is always the correct one since that one will be the product formed and the other reagent will react as needed and the remainder will be in excess and be left over.
So 1.25 mol CO2 will be formed.
grams CO2 = mols CO2 x molar mass CO2 = 1.25 x 44 = ?
Where did you get 125?
Nevermind my last question
To find the grams of CO2 produced in the reaction, we need to use stoichiometry and convert the given masses of CaCO3 and HCl to moles, calculate the mole ratio between CaCO3 and CO2, and then convert the obtained moles of CO2 back to grams.
First, let's calculate the number of moles of CaCO3:
Molar mass of CaCO3 = (40.08 g/mol for Ca) + (12.01 g/mol for C) + (3 x 16.00 g/mol for 3 O) = 100.09 g/mol
Moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3
Moles of CaCO3 = 125 g / 100.09 g/mol
Next, let's calculate the number of moles of HCl:
Molar mass of HCl = (1.01 g/mol for H) + (35.45 g/mol for Cl) = 36.46 g/mol
Moles of HCl = mass of HCl / molar mass of HCl
Moles of HCl = 125 g / 36.46 g/mol
Now, we need to determine the mole ratio between CaCO3 and CO2 based on the balanced equation:
1 mole of CaCO3 produces 1 mole of CO2
Since both the moles of CaCO3 and HCl are equal to each other, we can use either of them to calculate the moles of CO2.
Finally, let's convert the moles of CO2 to grams:
Molar mass of CO2 = (12.01 g/mol for C) + (2 x 16.00 g/mol for 2 O) = 44.01 g/mol
Grams of CO2 = moles of CO2 x molar mass of CO2
So, the grams of CO2 produced by reacting 125 g of CaCO3 with 125 grams of HCl can be calculated by the above steps.