Question

Complete the following equation
$CuS{O_4} + NaOH \to $

Answer 1

Hint: The compounds formed on the product side must follow the rules of conserved mass. That is the number of atoms formed on the product side must be equal to the number of atoms present in the reactant side. It is required to write the products in a balanced reaction.

Complete answer:
The above reaction shows the combining of copper sulphate and sodium hydroxide. To write the products that are being formed in this reaction, we have to first split the given reactants into their respective ions. This is demonstrated below:
$CuS{O_4} \to C{u^{ + 2}} + S{O_4}^{ - 2}$
$NaOH \to N{a^ + } + O{H^ - }$
Now we can form the products by combining the ions. This is done as follows:
$C{u^{ + 2}} + O{H^ - } \to Cu{\left( {OH} \right)_2}$
Since two hydroxide molecules are required we will add $2$ to the left hand side of the hydroxide molecule.
$C{u^{ + 2}} + 2O{H^ - } \to Cu{\left( {OH} \right)_2}$
Now for the sodium ion part,
$N{a^ + } + S{O_4}^{ - 2} \to N{a_2}S{O_4}$
We can see that the product side contains two atoms of sodium. Therefore we will balance the reaction on the sodium side as follows:
$2N{a^ + } + S{O_4}^{ - 2} \to N{a_2}S{O_4}$
In this way, the products formed are, $N{a_2}S{O_4}$ also known as sodium sulphate and the pther compound is $Cu{\left( {OH} \right)_2}$ which is also known as cuprous hydroxide.
Now we must combine the above equation to remove all the free ions. The new reaction will be represented as follows:
$CuS{O_4} + 2NaOH \to Cu{\left( {OH} \right)_2} + N{a_2}S{O_4}$
Therefore, the answer to the question will be
$CuS{O_4} + 2NaOH \to Cu{\left( {OH} \right)_2} \downarrow + N{a_2}S{O_4}$

Note:Remember that the equation should always be balanced. To carry out this, the atoms on the reactant side should be equal to the atoms on the product side. This is done to make sure that mass of the reaction is conserved as new mass cannot be created and mass cannot be destroyed.