0.3 kg of NaOH is dissolved in 5 litres of water. It is used as a titrant with 20.0cm3 of H2SO4 of a concentration of 0.5mol/dm3. what is the volum eof NaOH solution that is required to neutralize all the H2SO4?
For reference purposes, here's the equation for the reaction:
2 NaOH + H2SO4 -> 2 H2O + Na2SO4
Find concentration of NaOH = 300/5
= 60g/dm3
No. of moles of of H2SO4 = concentration X volume
=0.5 X 0.02
=0.001 moles
Using mole ratio, no of moles of NaOH =0.01 X 2
=0.02 moles
Now change 60g/dm3 to mol/dm3
concentration of NaOH in mol/dm3 = 60 / 40
=1.5mol/dm3
concentration of NaOH (dm3) = no.of moles/vol of NaOH used (manipulate with formula)
= 0.02/1.5
= 0.0133dm3 (answer)
2nd question.
5g of barium nitrate would react with 15cm3 of 0.3mol/dm3 of sodium sulphate. Determine total mass of products formed from reaction. (aint sure of how you do this. but i took a stab at the question anyway. correct me where necessary)
Equation of reaction = Ba(NO3)2 + Na2SO4 -> BaSO4 + 2NaNO3
find No. of moles of sodium sulphate
concentration of sodium sulphate (mol/dm3) = no.of mol/vol of solution
0.3 = no. of moles /0.015
manipulate with terms given and you get 0.0045 moles of Na2SO4.
No. of moles needed for BaSO4 (use mole ratio) : 0.0045moles
No of moles needed for NaNO3 = 0.0045 * 2
= 0.0090moles
Mr of BaSO4 = 233
Mr of NaNO4 = 85
No of moles (BaSO4) = mass/Mr
Mass = 233 * 0.0045
= 1.0485g
No of moles of NaNO3 = mass/mr
mass : 85 *0.0090 = 0.765g
total mass = 1.0485 +0.765
= 1.81g (ans)
3rd question.
Bromine liquid is produced when chlorine gas is pumped into a beaker filled with 100ml of 80g/dm3 of lithium bromide.
(a) find vol of chlorine gas needed to produce 4.5 g of bromine liquid.
(b) determine amount of LiCl produced when 2g of bromione liquid is also produced.
equation - 2 LiBr + Cl2 -> 2 LiCl + Br2
Find the no. of moles of bromine. (0.028moles)
compare using mole ratio.
(a)no. of chlorine moles needed is also 0.028moles.
0.027 = vol of gas(dm3) / 24dm3
vol. of Cl2 gas produced = 0.028 *24 = 0.672dm3(ans)
(b) Mr of Br2 = 80(2)
=160
No of moles of Br2 = mass/mr = 2/160 = 0.0125moles.
using mole ratio : 2*0.0125
= 0.025moles(ans)
lengthy post. thanks for going through the entire entry. please help identify mistakes wherever necessary.
Yes, all your answers are correct.
However, for the 2nd question, you neglected to determine limiting reactant, or you neglected to show the required working if you did (indeed determine limiting reactant). You appeared to simply assume sodium sulphate was the limiting reactant, and you just happened to assume the correct limiting reactant. In the exam you would lose some marks for that.
Similarly for the 3rd question. It is technically proper to check that the lithium bromide won't be a limiting problem (as it turned out, it isn't), instead of just assuming it wouldn't, as you did.
I might very well set such a question for my students in a test/exam, making lithium bromide limiting, in which case the correct answer would then be - "no volume of chlorine gas would do the job, since it's not possible to obtain 4.5 g of bromine liquid. The maximum obtainable mass/moles from would be..."
Otherwise... good job, bonkysleuth! *Hancock's thumbs up*