Question

Consider the reaction. 2 Al ( s ) + Fe 2 O 3 ( s ) heat −−→ Al 2 O 3 ( s ) + 2 Fe ( l ) If 17.3 kg Al reacts with an excess of Fe 2 O 3 , how many kilograms of Al 2 O 3 will be produced?

Answer 1

Answer:

32.7 kilograms of aluminium oxide  will be produced.

Explanation:

$2 Al ( s ) + Fe_2O_3 ( s ) +heat\rightarrow Al_2O_3 ( s ) + 2 Fe ( l )$

Mass of aluminum = 17.3 kg = 17300 g (1 kg = 1000 g )

Moles of aluminium = $\frac{17300 g}{27 g/mol}=640.74 mol$

According to reaction, 2 moles of aluminum gives 1 mole of aluminum oxide,then 640.74 moles of aluminum will give:

$\frac{1}{2}\times 640.74 mol= 320.37 mol$ of aluminum oxide

Mass of 320.37 moles of aluminum oxides:

320.37 mol × 102 g/mol = 32,677.74 g = 32.67774 kg ≈ 32.7 kg

32.7 kilograms of aluminium oxide  will be produced.