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Balance the following Acid/Base Reactions:3. __Cu(OH) + __H2CO3 --> __Cu2CO3 + _H2O

User Crollster
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To balance the equation it is necessary to have the same number of atoms of each element on both sides of the equation (reactant side and product side). Let's count the number of atoms of each element for the unbalanced equation:

__Cu(OH) + __H2CO3 --> __Cu2CO3 + _H2O

Reactant side:

Cu - 1

O - 4

H - 3

C - 1

Product side:

Cu - 2

O - 4

H - 2

C - 1

As it is an acid and base reaction, there is a trick to balance the reaction faster. We use the quantity of atoms of the cation of the salt on the products side as stoichiometric coefficient of the hydroxide and the quantity of the anion of the salt as stoichiometric coefficient of the acid.

In this case, the salt is:

Cu2CO3

So we have 2 cations and 1 anion.

So the coefficient of the hydroxide will be 2 and for the acid will be 1:

2 Cu(OH) + 1 H2CO3 --> __Cu2CO3 + _H2O

The coefficient of the salt will be 1 and for the water (H2O) is the coefficient of the acid multiplied by the coefficient of the acid (2x1 = 2):

2 Cu(OH) + 1 H2CO3 --> 1 Cu2CO3 + 2 H2O

To make sure that everything is right, let's count the number of atoms of each element for the balanced equation.

Reactant side:

Cu - 2

O - 5

H - 2

C - 1

Product side:

Cu - 2

O - 5

H - 2

C - 1

Now the equation is balanced.

Answer: 2 Cu(OH) + 1 H2CO3 --> 1 Cu2CO3 + 2 H2O

The coefficients are: 2, 1, 1, 2.

User Astra Bear
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