Question

If 54 g Al reacts with 160 g of O2. Find out mass of product (Al=27,O=16)​

Answer 1

number of atom n =mass ÷molar mass

number of Al atom =54÷27=2atom of Al.

number of o atom = 160÷16=10 atom of o .

molecules of ALO2

= 3×6.022×10 (23)

18.066×10 (23)

NOW ,MASS OF ALO2

MASS OF AL MOLECULES +MASS OF O2 MOLECULES

54+160

=214U MASS of ALO2.

Answer 2

Answer:

The weight of Aluminium oxide is 102 grams.

Explanation:

Given,

The weight of Aluminium (Al) W₁ = 54 g.

The weight of Oxygen(O₂) W₂ = 160 g.

To find,

The mass of the product formed i.e. Al₂O₃.

Calculation,

The reaction of Aluminium with oxygen gives aluminium oxide:

$4Al + 3O_2 \rightarrow 2Al_2O_3$

From the above reaction, we can see that 4 moles of aluminium react with 3 moles of oxygen.

But no. of moles of Aluminium(n₁) given is 54/27 moles = 2 moles.

And no. of moles of Oxygen (n2) given is 160/32 moles = 5 moles.

Hence, if 4 moles of aluminium react with 3 moles of oxygen, then 2 moles of aluminium react with 1.5 moles of oxygen.

So, Aluminium is the limiting reagent and oxygen is in excess.

Hence, if 4 moles of Aluminium react with 2 moles of Al₂O₃, then 2 moles of Aluminium react with:

$\frac{2 \times 2}{4}$ = n = 1 mole of Aluminium oxide.

Now we can calculate the mass of Aluminium oxide as:

weight of Aluminium oxide(w) = no. of moles of Aluminium oxide(n) × G.M.W of Aluminium oxide.

w = n × G.M.W

⇒ w = 1 × 102

Therefore, the weight of Aluminium oxide is 102 grams.

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