Need help. Sodium Sulphate crystallises out as Na2SO4.10H2O. It is prepared by titration. 25.0cm^3 aquas of NaOH solution having a concentration of 2.24 mol/dm^3 reacted with H2SO4 to give Na2SO4.10H2O. The molecular mass of Na2SO4.10H2O = 322 grams. Find the % yield of Na2SO4.10H2O.

Assuming the mass of Na2SO4.10H2O obtained from the crystallization was: 8.5g

The % yield is 94.3%

Explanation:

Assuming the mass of Na2SO4.10H2O was:

Based on the reaction:

2NaOH + H2SO4 + 8H2O →Na2SO4.10H2O

Where 2 moles of NaOH reacts producing 1 mole of Na2SO4.10H2O

To solve this question we need to find the moles of NaOH added. With the moles and the reaction we can find the moles of Na2SO4.10H2O and its actual theoretical yield. Assuming the mass of Na2SO4.10H2O produced was: 8.5g

we can find the percent yield as follows:

Moles NaOH:

25.0cm³ = 0.0250dm³ * (2.24mol/dm³) = 0.056 moles NaOH

Moles Na2SO4.10H2O:

0.056 moles NaOH * (1mol Na2SO4.10H2O / 2mol NaOH) = 0.028 moles Na2SO4.10H2O

Mass Na2SO4.10H2O:

0.028 moles Na2SO4.10H2O * (322g / mol) = 9.016g Na2SO4.10H2O = Theoretical yield

Percent yield is 100 times the ratio between actual yield:

8.5g / 9.016g * 100 = 94.3%