Question

A mass of `2.40*g` of calcium oxide is obtained by heating a mass of `5.00*g` calcium carbonate. What is the stoichiometric equation that represents the formation of calcium oxide, and what is the percentage yield?

Answer 1

We need a stoichiometrically balanced equation to represent the decomposition of calcium carbonate..........

Explanation:

`CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)uarr`

The `Delta` symbol represents heat, and you have to fiercely heat these carbonates to effect decomposition. Note that this reaction is certainly stoichiometrically balanced: garbage in equals garbage out.

`"Moles of calcium carbonate "=" "(5.00*g)/(100.09*g*mol^-1)`

`=` `0.0500*mol`

`"Moles of calcium oxide "=" "(2.40*g)/(56.08*g*mol^-1)`

`=` `0.0428*mol`

`"% yield"` `=` `(0.0428*mol)/(0.0500*mol)xx100%=86%`

Are you happy with this?

Note that most carbonates behave this way under heat:

`MCO_3(s) + Delta rarr MO(s) + CO_2(g)uarr`