Consider the following reaction: Al + HBr ---> AlBr3 + H2 . When 4 moles of Al reacts with 3 moles of HBr, how many moles of H2 are formed?

2Al + 6HBr ==> 2AlBr3 + 3H2

4 mols Al will produce how much H2? That's 4 mols Al x (3 mols H2/2) mols Al) = ?

3 mols HBr will produce how much H2?That's 3 mols HBr x (3 mols H2/6 mols HBr) = ?

This is a limiting reagent problem. The answer to how much is produced is ALWAYS the smaller value of the two possibilities.

To find the number of moles of H2 formed when 4 moles of Al reacts with 3 moles of HBr, we can use the stoichiometric ratio given in the balanced chemical equation.

The balanced chemical equation is:
Al + HBr ---> AlBr3 + H2

From the balanced equation, we can see that the stoichiometric ratio between Al and H2 is 1:1. This means that for every 1 mole of Al that reacts, 1 mole of H2 is formed.

Therefore, if 4 moles of Al react, we can conclude that 4 moles of H2 will be formed.

According to the balanced chemical equation for the reaction:

Al + 3HBr ---> AlBr3 + 3H2

The stoichiometry of the balanced equation tells us that for every 3 moles of HBr reacting, 3 moles of H2 are produced.

Therefore, if 3 moles of HBr produced 3 moles of H2, we can use this ratio to determine the number of moles of H2 formed when 4 moles of Al reacts with 3 moles of HBr.

Using the ratio, we can set up a proportion:

3 moles of HBr / 3 moles of H2 = 4 moles of HBr / x moles of H2

Cross-multiplying, we have:

3 moles of HBr * x moles of H2 = 4 moles of HBr * 3 moles of H2

This simplifies to:

3x moles of H2 = 12 moles of H2

Dividing both sides of the equation by 3, we find:

x moles of H2 = 12 moles of H2 / 3

Therefore, when 4 moles of Al reacts with 3 moles of HBr, 4 moles of HBr are formed.