How many grams of Br2 are needed to completely convert 15.0g Al to Albr3?

For, this question,

The balanced equation is; 2Al + 3 > 2Al

We can now, calculate how many grams of will be needed to completely convert 15 g of Al into Al.

Here, the reaction produces, 2 moles of Al

So, For 15.0 g of Al; we can calculate;

(15.0 g Al) X (1 mole Al / 26.98 g Al) X (3 moles / 2 moles Al) X (159.81 g / 1 mole )

= 133 g of  will be needed.

Therefore, 133g of will be needed to completely convert 15 g of Al into Al.

Explanation:

може ли да зареждате повече информация, моля