Question

How many moles of Au and grams of Au are present in 500g of Au₂O₃? (Au= 19)

Answer 1

SOLUTION : Mol. wt. of $Au _{2} O _{3}=197 \times 2+16 \times 3=442$

Now, $500 g$ of $Au _{2} O _{3}=\frac{500}{442}$ moles of $Au _{2} O _{3}=1 \cdot 13$ moles.

From formula it is clear that one mole of $Au _{2} O _{3}$ contains 2 moles of $Au$ $\therefore 1 \cdot 13$ moles of $Au _{2} O _{3}$ contain $2 \times 1.13$ moles $= 2 \cdot 2 6$ moles of $A u$.

Ağain, 1 mole of $A u=197 g$

$\therefore 2.26$ moles of $A u=(197 \times 2 \cdot 26) g = 4 4 5 \cdot 2 g$.