and please if somebody can do this...
OK, hopefully this clears your doubt.
This time, you'll need to also apply this formula:
Concentration (mol/dm
3) = Moles / Volume (dm
3)
So:
i) Number of moles of NaOH used = 2.24*0.025 =
0.056 molsii) Maximum number of moles of Na
2SO
4.10H
2O that could be formed = From the equation given, you can see 2 moles of NaOH gives 1 mole of Na
2SO
4.10H
2O. So it is simply: 0.056/2 =
0.028 molsiii) Mass of one mole of Na
2SO
4.10H
2O = 322g
iv) Maximum yield of sodium sulphate-10-water = 0.028*322 =
9.02gv) Percentage yield = (3.86/9.02)*100 =
42.8%If you don't understand any of this, I'll be happy to explain