Question

If `171*g` `Ba(OH)_2` are reacted with `294*g` of `H_2SO_4`, which is the reagent in excess? How much `"barium sulfate"` will be deposited?

Answer 1

What 5 steps......?

Explanation:

First, we write a stoichiometrically balanced equation......

`Ba(OH)_2(s) + H_2SO_4(aq)rarrBaSO_4(s)darr+2H_2O`

The equation is stoichiometrically balanced, and shows a 1:1 molar equivalence between barium hydroxide and sulfuric acid.

And then we calculate the molar equivalents of each starting reagent:

`"Moles of"` `BaOH_2=(171*g)/(171.34*g*mol^-1)=1*mol`

`"Moles of"` `H_2SO_4=(294*g)/(98.08*g*mol^-1)=3*mol`

You can see that they are spoon-feeding you a bit in the question, given the molar equivalence. Clearly, there IS EXCESS sulphuric acid. Do you agree?

Given that there is excess sulfuric acid, clearly, the reaction will be stoichiometric in barium hydroxide, and ONE MOLE of barium sulfate will be generated, i.e. `233.43*g`. And there will be excess sulfuric acid, to the tune of `2*mol`, i.e. `196*g`.

And you will have to adapt your given 5-step recipe to this treatment. The most important step was writing the chemical equation, which establishes absolute molar (and mass) equivalence.

Note that this question is a little bit abstract. In the lab we do not measure out sulfuric acid by mass. We would take (say) `1*mol*L^-1` solution of the acid, and then add this to the barium salt. How many millilitres of `1.0*mol*L^-1` `H_2SO_4(aq)` would be required for molar equivalence in this reaction?