2Al(s) + 6 Hcl(aq) ----- 2AlCl3(aq) + 3H2(g). step by step?

Question

how to balance the equation step by step thank you

Answer 1

Al(s) +  HCl(aq) --> AlCl3(aq) + H2(g)

Start by making a list of the elements and the number of each on both sides of the arrow.  

Al(s) +  HCl(aq) --> AlCl3(aq) + H2(g)

Al 1 ... .... ... ... ... ... .. Al  1

H  1... .... ... ... ... ... ... H  2

Cl  1 ... .... ... ... ... ... ..Cl 3

Now, the idea is to start at the top of the list and change a each coefficient as you go down the list.  Each time you change a coefficient, you must run a new set of numbers.  Repeat till  the number of each element is the same.  It's ok if you unbalance an previous element, it will get fixed later.  When you get to the bottom of the list and its still not balanced, start back over at the top of the list and come down again.

The aluminum is balanced, so place a 2 in front of HCl, and run a new list on that side.

Al(s) +  2HCl(aq) --> AlCl3(aq) + H2(g)

Al 1 1... .... ... ... ... ... .. Al  1

H  1.2.. .... ... ... ... ... ... H  2

Cl  1 2... .... ... ... ... ... ..Cl 3

Now we see that we need to change the 2 in front of HCl to a 3 to balance the Cl.  It's Ok to change a coefficient.

Al(s) +  3HCl(aq) --> AlCl3(aq) + H2(g)

Al 1 1..1. ...... ... ... ... .. Al  1

H  1.2..3... ... ... ... ... ... H  2

Cl  1 2..3 ... ... ... ... ... ..Cl 3

Now you must recognize that there is an impasse and look for a number of H that 2 and 3 will both go into.  That would be six.  Write two coefficients (one on each side) so that H will have 6 atoms on each side.

Al(s) +  6HCl(aq) --> AlCl3(aq) + 3H2(g)

Al 1 1..1.1 ...... ... ... ... .. Al  1 1

H  1.2..3..6. ... ... ... ... ... H  2  6

Cl  1 2..3 .6.. ... ... ... ... ..Cl 3  3

Now double the Cl on the right side by putting  2 in front of AlCl3 and run a new set of numbers.  Remember.  Each time you change a coefficient, you must run a new set of numbers.

Al(s) +  6HCl(aq) --> 2AlCl3(aq) + 3H2(g)

Al 1 1..1.1 ...... ... ... ... .. Al  1 1 2

H  1.2..3..6. ... ... ... ... ... H  2  6 6

Cl  1 2..3 .6.. ... ... ... ... ..Cl 3  3 6

Now go back to the top.  Balance the Al and you're done.  You will notice that the two rightmost columns on each side of the arrow match, telling you that the equation is balanced.

2Al(s) +  6HCl(aq) --> 2AlCl3(aq) + 3H2(g)

Al 1 1..1.1 2.... ... ... ... .. Al  1 1 2

H  1.2..3..6. 6... ... ... ... ... H  2  6 6

Cl  1 2..3 .6..6 ... ... ... ... ..Cl 3  3 6

Answer 2

It is already balanced...

But suppose you have Al (s) + HCl (aq) ------ AlCl3 (aq) + 3 H2 (g)

1. Al (s) + 6 HCl (aq) ------ 2 AlCl3 (aq) + H2 (g)

Consider the subscripts first. Since Cl has 3 the other side should also have 3, but, putting 3 there would have a conflict with H2 which has 2 atoms on its subscript. Think of a number that would make it equal for both... 6! Since you put 2 in the product side for Al Cl3, you should also put 2 in the reactant side...

2. 2Al (s) + 6HCl (aq) ------ 2AlCl3 (aq) + H2 (g)

3. 2Al (s) + 6HCl (aq) ------ 2AlCl3 (aq) + 3H2 (g)

Answer 3

1)  2 Al on left hand side, 2 Al on right hand side

2)  Now, 3 x 2 = 6 Cl on right , therefor 6 Cl on left hand side

3)  Because H on right has 2 already: put 3 so that 3x2=6 on right hand side and 6 H on left = 6 H on right