CAIE IGCSE Chemistry 3.3.8 Percentage yield, percentage

Practise percentage yield, composition and purity calculations from equations, masses and formulae in common inorganic reactions.

Exam points

calculate percentage yield from actual mass and theoretical mass from the equation
find percentage composition by mass using Mr and the mass of the named element
use percentage composition data to compare ores or compounds as better extraction sources

[Maximum number: 3]

Sulfur is an important element.

Iron(II) sulfate decomposes when heated strongly.

2 \mathrm{FeSO}_{4}(\mathrm{~s}) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{SO}_{2}(\mathrm{~g})+\mathrm{SO}_{3}(\mathrm{~g})

15.20 g of $\mathrm{FeSO}_{4}(\mathrm{~s})$ was heated and formed 4.80 g of $\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})$ .
$\left[M_{\mathrm{r}}, \mathrm{FeSO}_{4}=152 ; M_{\mathrm{r}}, \mathrm{Fe}_{2} \mathrm{O}_{3}=160\right]$
Calculate the percentage yield for this reaction.
[Total: 26]

$\mathrm{M} 1 \mathrm{Mol} \mathrm{FeSO} \mathrm{L}_{4}=15.2 / 152=0.1(00)$

M2 Expected mol of $\mathrm{Fe}_{2} \mathrm{O}_{3}=0.1 / 2=0.05(00)$ )

or

Actual mol of $\mathrm{Fe}_{2} \mathrm{O}_{3}=4.80 / 160=0.03(00)$

M3 Percentage yield $=100 \times 0.03(00) / 0.05(00)=60 \%$

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3.3.8—Percentage yield, percentage question 1