Question

Directions: show all your work, box your answer, and use significant figures to receive credit.What mass of oxygen is needed to react with 55.1g of aluminum in the synthesis of aluminum oxide?

Answer 1

49.0 grams of oxygen is needed.

Step-by-step explanation

Given:

Mass of Aluminum = 55.1 g

What to find:

The mass of oxygen needed to react with the Al.

Step-by-step solution:

The equation for the reaction is:

4Al + 3O₂ → 2Al₂O₃

From the equation; 4 moles of Al reacts with 3 moles of O₂

1 mole of Al = 26.982 g/mol

1 mole of O₂ = 31.998 g/mol

It implies;

(4 mol x 26.982 g/mol) = 107.928 g of Al reacts with (3 mol x 31.998 g/mol) = 95.994 g of O₂

So, 55.1 g of Al will need

`\frac{55.1\text{ }g\text{ }Al}{107.928\text{ }g\text{ }Al}*95.994\text{ }g\text{ }O_2=49.0\text{ }g\text{ }O_2`

Therefore, the mass of oxygen needed to react with 55.1g of aluminum in the synthesis of aluminum oxide is 49.0 grams