Question

Aluminum reacts with oxygen. What is the balanced equation? If 1.295 g of aluminum reacts with 0.48 g of oxygen, how many grams of aluminum oxide are formed? Identify the limiting reactant. Include balanced equation, molar mass values, dimensional analysis, and limiting reactant.

Answer 1

$4Al + 3O_2 → 2Al_2O_3$

This is the balanced equation.

We have given that,

1.295g of Aluminum reacts with 0.48g of Oxygen

 No. of moles of Al = $\dfrac{1.295}{27}$ = 0.048

No. of moles of oxygen=$\dfrac{0.99}{32}$ = 0.031

For 4 moles of Al we need 3 moles of $O_2$

0.048 moles of Al we need 0.03 moles of $O_2$

Therefore some oxygen is left behind

Hence, Aluminum is limiting reagent.