Question

Question #2b66a

Answer 1

`2NaOH(aq) + H_2SO_4(aq) rarr Na_2SO_4(aq) + 2H_2O(l)`

Explanation:

The given equation is absolutely vital to answering this question. It says that 2 mol sodium hydroxide, approx. `80.0*g`, react with the 1 mol sulfuric acid, a mass of approx. `98.08*g`, to give stoichiometric quantities of sodium sulfate and water.

So we must calculate the molar quantities of our starting materials.

`"Moles of sodium hydroxide"=(20.0*g)/(40.00*g*mol^-1)` `=` `0.5*mol`.

`"Moles of sulfuric acid"=(9.8*g)/(98.08*g*mol^-1)` `=` `0.1*mol`.

So, clearly, sulfuric acid is the limiting reagent. And at most, I can get `0.2*mol` of `NaOH` to react.

Given this, only `0.1*mol` of `Na_2SO_4` will be formed, i.e. `0.1*molxx142.04*g*mol^-1` `=` `14.2*g`.

All of these calculations relies on the given stoichiometric reaction. If I don't have this, or if I have not balanced it correctly, I will not be able to do the calculation. Capisce?