neutralisation?

Question

If 49g of H2SO4 is dissolved and then neutralised by some NaOH producing 63g of Na2SO4 how do I work out what the yield of Na2SO4 is?

do I need to work out the mass and then the moles of each substance or is there a simpler way of working it out?

Thanks

Answer 1

First you need the equation for the reaction.

2NaOH + H2SO4 = Na2SO4 + 2H2O

This shows that 1 mole of acid will make 1 mole of Na2SO4

1 mole of H2SO4 = 98g ( 2 H = 2 ; S = 32 ; 4 O = 64)

1 mole Na2SO4 = 142g ( 2 Na = 46 S = 32 4 O = 64)

So 49g is half a mole of acid and should make 71g of sodium sulphate.

As only 63g was made the yeild is 63 divided by 71 and multiplied by 100 to make it into a percentage. = 88.73%

Answer 2

H2SO4 + 2NaOH = Na2SO4 + 2H2O
1 : 2 :: 1 : 2 Molar Ratios

From the formula it will be noticed that one molar ratio of sulphuric acid reacts to form one molar ratio of sodium sulphate.

Moles (H2SO4) = 49g/ 98 Mr(H2SO4) = 0.5 moles
Moles (Na2SO4) = 0.5 moles
because the molar ratios are the same.

Theoretical yield of Sodium Sulphate/g = 0.5 moles x 142 Mr (Na2SO4).
= 71g
However, the actual yield is 63g.
Therefore percentage yield is 63/71 x 100 = 88.73%

Answer 3

Since 1 mole of the acid (98 grms) makes 1 mole of salt (142 grs)
then 49 grms should make 71 grms of the salt.
If you only have 63 grm then you have 63/71 (fraction) of what you could have or 100*63/71 as a percent.
88.7%

Answer 4

need to use a balanced equation firstly. then convert grams to moles then work around it and then back to grams