- #1

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1)H2SO4 + Al(OH)3 --> Al2(SO4)3 + H2O

2)Pb2+ + Cl- --> PbCl2

and word equation like this

ethane(C2H6) + Oxygen --> Carbon dioxide + Water vapour

Thq.

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- Thread starter milkyman
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- #1

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1)H2SO4 + Al(OH)3 --> Al2(SO4)3 + H2O

2)Pb2+ + Cl- --> PbCl2

and word equation like this

ethane(C2H6) + Oxygen --> Carbon dioxide + Water vapour

Thq.

- #2

- 907

- 2

Balancing chemical equations follows the rule of conservation of mass. Thus in the chemical equations, the quantity of a specific element on the reactants side must be equal to the quantity of the same specific element on the product side.

I'll do one for you and then hopefully you can figure the rest out.

In the chemical reaction [tex]\epsilon H_2SO_4 + \epsilon Al(OH)_3 \rightarrow \epsilon Al_2(SO_4)_3 + \epsilon H_2O[/tex] we have [tex]\epsilon[/tex] of each element on each side and they must be equal to each other (our goal is to find out what the epsilons are, they can be different; [tex]\{\epsilon | \epsilon \in \mathbb{R}\}[/tex]) thus

[tex]\epsilon H_{(r)}=\epsilon H_{(p)},

\epsilon S_{(r)}=\epsilon S_{(p)},

\epsilon O_{(r)}=\epsilon O_{(p)},

\epsilon Al_{(r)} = \epsilon Al_{(p)}

[/tex]

(The (r), and (p) indices just stand for reactant and product)

substituting the amounts in we find that

[tex]5H \neq 2H,

S \neq 3S,

7O \neq 13O,

Al \neq 2Al[/tex]

It is then a priority to make all of these inequalities ([tex]\neq[/tex]) equalities which is what balancing is.

By adding co-efficients to the chemicals/formula units/elements in the reaction we end up with the equation [tex]3H_2SO_4 + 2Al(OH)_3 \rightarrow Al_2(SO_4)_3 + 6H_2O[/tex], this may not be immediate obvious why this is but if you set up a table and play around with the numbers you'll eventually get a viable solution. therefore:

[tex]12H_{(r)} = 12H_{(p)},

3S_{(r)} = 3S_{(p)},

18O_{(r)} = 18O_{(p)},

2Al_{(r)} = 2Al_{(p)}[/tex]

In terms of the word equations, [tex]Ethane\ +\ Oxygen \rightarrow Carbon\ Dioxide + Water\ Vapour[/tex] it can be rewritten as [tex]C_2H_6 + O_2 \rightarrow CO_2 + H_2O[/tex], this is a combustion reaction. Follow the same procedure as above to balance it.

I'll do one for you and then hopefully you can figure the rest out.

In the chemical reaction [tex]\epsilon H_2SO_4 + \epsilon Al(OH)_3 \rightarrow \epsilon Al_2(SO_4)_3 + \epsilon H_2O[/tex] we have [tex]\epsilon[/tex] of each element on each side and they must be equal to each other (our goal is to find out what the epsilons are, they can be different; [tex]\{\epsilon | \epsilon \in \mathbb{R}\}[/tex]) thus

[tex]\epsilon H_{(r)}=\epsilon H_{(p)},

\epsilon S_{(r)}=\epsilon S_{(p)},

\epsilon O_{(r)}=\epsilon O_{(p)},

\epsilon Al_{(r)} = \epsilon Al_{(p)}

[/tex]

(The (r), and (p) indices just stand for reactant and product)

substituting the amounts in we find that

[tex]5H \neq 2H,

S \neq 3S,

7O \neq 13O,

Al \neq 2Al[/tex]

It is then a priority to make all of these inequalities ([tex]\neq[/tex]) equalities which is what balancing is.

By adding co-efficients to the chemicals/formula units/elements in the reaction we end up with the equation [tex]3H_2SO_4 + 2Al(OH)_3 \rightarrow Al_2(SO_4)_3 + 6H_2O[/tex], this may not be immediate obvious why this is but if you set up a table and play around with the numbers you'll eventually get a viable solution. therefore:

[tex]12H_{(r)} = 12H_{(p)},

3S_{(r)} = 3S_{(p)},

18O_{(r)} = 18O_{(p)},

2Al_{(r)} = 2Al_{(p)}[/tex]

In terms of the word equations, [tex]Ethane\ +\ Oxygen \rightarrow Carbon\ Dioxide + Water\ Vapour[/tex] it can be rewritten as [tex]C_2H_6 + O_2 \rightarrow CO_2 + H_2O[/tex], this is a combustion reaction. Follow the same procedure as above to balance it.

Last edited:

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- #4

symbolipoint

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For your aluminum sulfate one, notice the formula: Al

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- #6

Borek

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In the chemical reaction [tex]\epsilon H_2SO_4 + \epsilon Al(OH)_3 \rightarrow \epsilon Al_2(SO_4)_3 + \epsilon H_2O[/tex] we have [tex]\epsilon[/tex] of each element on each side and they must be equal to each other

Lousy use of symbols IMHO. While you have explained [tex]\epsilon[/tex] are different, they look to be the same, that's not helping in understanding what is going on.

- #7

Borek

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In the case of really difficult reaction equations you can try algebraic method of balancing reaction equations, which means solving system of equations, and as such is not a trial and error but can be done in a systematic way. But for most cases it is an overkill and is in no way simpler/faster than the inspection method. Inspection in most cases works nice if you follow simple rules outlined on the linked page.

Best you can do is to balance as many equations as you can, sooner or later you will be balancing them on the fly.

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- #9

symbolipoint

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