Question

how many moles of al2o3 will be formed when micture of 5.4 g of al and 3.2 g of o2 is heated

Answer 1

Answer:

6.8 gram

Explanation:

We need balanced chemical equation for the given reaction.

The balanced chemical equation of given reaction is :-

$\sf 4Al + 3O_2 \rightarrow 2Al_2O_3$

Now, from the balanced equation we can say that 4 moles of Al requires 3 moles of Oxygen gas to produce 2 moles of Al2O3.

Let's find out the given moles of each element.

mole = given mass/molar mass

→ mole of Al = 5.4/27

→ mole of Al = 0.2 moles

Mole of O2 = 3.2/32

→ mole of O2 = 0.1 moles

Now, since 3 moles of O2 burns with 4 moles of Al

→ 0.1 moles of O2 will burn with 4/3 × 0.1 = 0.4/3 moles of Al

Here, we have 0.3 moles of Al. So Aluminium is in excess. Only 0.4/3 moles will be burnt and rest will be left unreacted.

Now, 4 moles of Al gives 2 moles of Al2O3

→ 0.4/3 moles of Al will give 0.4/6 moles of Al2O3

Now, molar mass of Al2O3 = 102

Moles = given mass/molar mass

→ given mass = molar mass × moles

→ given mass = 102 × 0.4/6

→ mass = 6.8 gram

Hence, 6.8 grams of Al2O3 will be produced in given reaction.