How many formula units of CaCl2 are expected from 5.3g of calcium and excess chlorine?
When calcium reacts with chlorine, calcium chloride is formed. The balanced chemical reaction of this equation is `Ca + Cl_2 -> CaCl_2` .
One mole of calcium reacts with one mole of chlorine gas to give one mole of calcium chloride. One mole of a substance contains 6.023*10^23 formula units. The molar mass of calcium is 40 g/mole. 5.3 g of calcium contains `5.3/40 ~~` 0.1325 moles of calcium.
This is equivalent to 0.1325*6.023*10^23 ~~ 7.98*10^22 atoms of calcium. For each atom of calcium one formula unit of calcium chloride is created when it reacts with excess chlorine gas. This gives the number of formula units of calcium chloride created when 5.3 g of calcium reacts with excess chlorine as 7.98*10^22.
Calcium Chloride can be prepared by the reaction of metallic calcium with elemental chlorine as per the equation:
`Ca(s) + Cl_2(g) rarr CaCl_2(s)`
Molar mass of Calcium is 40.08 g/mol. Use of 5.3 g Ca and excess chlorine makes calcium the limiting reagent, hence the amount of product `(CaCl_2)` formed will depend upon the amount of calcium only.
From the balanced chemical equation of its formation, it is evident that,
40.08 g Ca, on reaction with excess chlorine, produces 1 formula unit of `CaCl_2` .
Therefore, 5.3 g Ca, on reaction with excess chlorine, should produce (1*5.3)/40.08, i.e. `0.132236~~0.132` formula units of `CaCl_2` .