And it was conveniently the same in this problem, too.
CuSO4 + 2NaOH ==> Cu(OH)2 + Na2SO4
You had 0.3 x 0.05 = 0.015 mols CuSO4.
You had 0.6 x 0.05 = 0.030 mols NaOH but you can see from the stoichiometry that it takes twice as much NaOH as CuSO4. You had only 0.015 mols Cu(OH)2 formed.
Sometimes the reagents DON'T "exactly" match in which case it is just another limiting reagent problem and something will be in excess.
As to why you added 50 mL + 50 mL you need to remember why you do a problem like this. It starts out to find dH for the reaction but you need some method by which to measure it. Therefore, you stick the reaction in a calorimeter and measure the heat produced BY MEASURING HOW MUCH THE WATER IS HEATED. You had 100 mL H2O so you use 100. It makes little difference to the chemicals that 50 came from one and 50 from the other; the idea is that you heated 100 mL H2O, you know the specific heat H2O, and you know the rise in temperature. Therefore, you can calculate the enthalpy change for the water which is (voila) the enthalpy change for the reaction.
50 mL of copper(II) sulfate reacts with 50 mL of sodium hydroxide. Their concentrations are 0.3 M and 0.6 M respectively. The temperature increased to 23.6 C from 23.4 C.
Determine the enthalpy change for the reaction in kJ/mol of sodium hydroxide.
Q=mcΔT
=(50g + 50g)(4.184 J/g C)(3.2 C)
= 1338.9 J
n=c x v
=0.600 mol/L x 0.050 L
=0.030 mol NaOH
ΔH = -Q/n
= -1.3339 kJ/0.030 mol
= -44.6 kJ/mol
What if it asked for the enthalpy change of the reaction (not specifically just in kJ/mol of sodium hydroxide). When you calculate for moles, would you add the moles of NaOH and CuSO4?
And you know how when you do calculations for the enthalpy change of a reaction you start sort of like this?:
NH3(aq) + HCl(aq) → NH4Cl(aq)
25 mL 25 mL
1.0 mol/L 1.0 mol/L
n = c x v
= 1 mol/L x 0.025L
= 0.025 mol
How come you only need the moles of one of the reactants? Because when we do Q=mcΔT, for m, we include the mass of both solutions... And for example in the first problem I showed, if the volume and concentration of the reactants are different, I don't think it'd be correct to just use the moles of one of the reactants to calculate ΔH for whole reaction...
For other sample problems, the concentrations and volumes of both reactants were conveniently the same. So we never had to choose which reactant to get the moles for (except in the problem I had at the beginning - but that one specifically asks for it in kJ/mol of NaOH). But what if the concentrations and volumes were different and we were never specifically asked to find it in kJ/mol of ,e.g., NaOH? Would that ever happen? Or maybe that's too advanced or something for right now? I'm just scared something like that will pop up on a test and I'm just going to have to guess which reactant to use to find moles...
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