Question

Question #99100

Answer 1

Consider the reaction,

`H_2SO_4(aq) + MgO(s) to MgSO_4(aq) + H_2O(l)`

Note: clearly this is a rough approximation of the reaction in the lab. Much more is happening (or not happening) in reality. Your question is relatively trivial so the preceding representation will suffice.

Hence,

`25.0cm^3 * ((dm)/(10cm))^3 * (1.0"mol")/(dm^3) approx 2.5*10^-2"mol"` of `H_2SO_4`

will result in (theoretically),

`2.5*10^-2"mol" * (MgSO_4)/(H_2SO_4) approx 2.5*10^-2"mol"` of `MgSO_4`

We obtain when the reaction is "completed",

`2.5*10^-2"mol" * (120.4g)/("mol") approx 3.01g`

Hence,

`(1.23g)/(3.01g) approx 40.9%`