DrBob222 I am beginning to understand about balancing equations well, thanks to your help, but there are still some other things, which I am having a hard time figuring out.

Complete and balance the following chemical reactions between an acid and a metal.

A) HBR + Mg ->

This is confusing for me, how do I balance this out?

There was a similar example like this in the book and it was ....

HBr + Al =

6HBr + 2Al -> 2AlBr3 + 3H2.

My questions are that how did they get "6" Hbr? Also where did the 2Al come from, For the product part, I know that the three comes from Aluminum, but why do they not add it in the first part. This is really confusing me now, I think I'm forgetting this whole balancing stuff. DrBob222 please bring me back on track.

A) HBR + Mg ->

This is confusing for me, how do I balance this out?
First, write the products.
HBr + Mg ==> MgBr2 + H2
Next, balance with coefficients.
2HBr + Mg ==> MgBr2 + H2
Finally, check it.
I see 2 H on both sides.
I see 2 Br on both sides.
I see 1 Mg on both sides.
Good. Balanced.


There was a similar example like this in the book and it was ....

HBr + Al =
Write the produces.
HBr + Al ==> AlBr3 + H2
Balance with coefficients.
I note quickly that I need a 2HBr to balance H on the right but a 3HBr to balance Br on the right. When that happens, we know neither 3 nor 2 will work so we go with 6.
6HBr + 2Al ==> 2AlBr3 + 3H2.
Finally, check it.
I see 6 H on left and right.
I see 6 Br on left and right.
I see 2 Al on left and right.
Good. Balanced.


6HBr + 2Al -> 2AlBr3 + 3H2.

My questions are that how did they get "6" Hbr? Also where did the 2Al come from, For the product part, I know that the three comes from Aluminum, but why do they not add it in the first part. This is really confusing me now, I think I'm forgetting this whole balancing stuff. DrBob222 please bring me back on track.
The 2 for 2AlBr3 is used because that's the only way to get 6 Br atoms and that forces us to place a 2 for Al on the left.

No problem, I'm here to help! Balancing chemical equations can be a bit tricky at first, but with practice, you'll get the hang of it.

Let's start by balancing the equation A) HBr + Mg -> ?

To balance this equation, we need to make sure that the number of each type of atom is the same on both sides of the equation. We'll go through it step by step.

1. Remember that in a chemical reaction, the total charge on both sides must be equal. In this case, both HBr and Mg are neutral, so we don't have to worry about charges.

2. Start by counting the number of atoms on each side of the equation. On the left side, we have 1 hydrogen (H) atom and 1 bromine (Br) atom. On the right side, we have 1 magnesium (Mg) atom.

Now, we need to balance the hydrogen and bromine atoms. Since hydrogen is already balanced, we'll focus on balancing the bromine. To do that, we'll need to add a coefficient in front of HBr. We'll use a coefficient of 2 to achieve that.

The equation becomes: 2HBr + Mg -> ?

Now, we have 2 hydrogen (H) atoms and 2 bromine (Br) atoms on the left side. But on the right side, we still have only 1 magnesium (Mg) atom. To balance this, we'll need to add a coefficient of 2 in front of Mg.

The final balanced equation is: 2HBr + 2Mg -> ?

So, the balanced equation for A) HBr + Mg -> ? is 2HBr + 2Mg -> 2MgBr + H2.

Now, let's address your second question regarding the example from the book: HBr + Al -> 6HBr + 2AlBr3 + 3H2.

In this example, the coefficient of 6 in front of HBr is used to balance the hydrogen (H) atoms. Additionally, the coefficient of 2 in front of Al is used to balance the aluminum (Al) atoms. The number 2 in AlBr3 was obtained by multiplying the coefficient of Al by 3 because there are three Bromine (Br) atoms in AlBr3. Don't forget that the coefficients must be the smallest whole numbers possible, so you can divide them if necessary.

Remember, balancing equations takes practice, so don't worry if it seems confusing at first. Just keep practicing, and it will become easier with time.