Taking into account the reaction stoichiometry, 5.23 grams of CaCl₂ is required to react with the 5.0 grams of sodium carbonate.
Reaction stoichiometry
The balanced reaction is:
CaCl₂ + Na₂CO₃ → CaCO₃ + 2 NaCl
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CaCl₂: 1 mole
- Na₂CO₃: 1 mole
- CaCO₃: 1 mole
- NaCl: 2 moles
The molar mass of the compounds is:
- CaCl₂: 110.9 g/mole
- Na₂CO₃: 106 g/mole
- CaCO₃: 100 g/mole
- NaCl: 58.45 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- CaCl₂: 1 mole ×110.9 g/mole= 110.9 grams
- Na₂CO₃: 1 mole ×106 g/mole= 106 grams
- CaCO₃: 1 mole ×100 g/mole= 100 grams
- NaCl: 2 moles ×58.45 g/mole= 116.9 grams
Rule of three
The rule of three is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.
If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied.
To solve a direct rule of three, the following formula must be followed, being a, b and c known data and x the variable to be calculated:
a ⇒ b
c ⇒ x
So: x=(c×b)÷a
Mass of CaCl₂ required
The following rule of three can be applied: If by reaction stoichiometry 106 grams of Na₂CO₃ react with 110.9 grams of CaCl₂, 5 grams of Na₂CO₃ react with how much mass of CaCl₂?
mass of CaCl₂= (5 grams of Na₂CO₃ ×110.9 grams of CaCl₂)÷ 106 grams of Na₂CO₃
mass of CaCl₂= 5.23 grams
Finally, 5.23 grams of CaCl₂ is required to react with the 5.0 grams of sodium carbonate.