# Question #54977

Apr 26, 2016

${\text{Cu"_ ((s)) + 2"H"_ 2"SO"_ (4(aq)) -> "CuSO"_ (4(aq)) + 2"H"_ 2"O"_ ((l)) + "SO}}_{2 \left(g\right)}$

#### Explanation:

The first thing to do here is write out the chemical formulas for the reactants and for the products.

• copper, $\text{Cu}$
• sulfuric acid, ${\text{H"_2"SO}}_{4}$

and your products are

• copper(II) sulfate, ${\text{CuSO}}_{4}$
• water, $\text{H"_2"O}$
• sulfur dioxide, ${\text{SO}}_{2}$

Write them out in the form of an unbalanced chemical equation

${\text{Cu"_ ((s)) + "H"_ 2"SO"_ (4(aq)) -> "CuSO"_ (4(aq)) + "H"_ 2"O"_ ((l)) + "SO}}_{2 \left(g\right)}$

In order to balance this chemical equation, you must make sure that all the atoms present on the reactant's side are also present on the products' side.

Notice that the reactants' side contains

• one atom of copper, $1 \times \text{Cu}$
• two atoms of hydrogen, $2 \times \text{H}$
• one atom of sulfur, $1 \times \text{S}$
• four atoms of oxygen, $4 \times \text{O}$

On the other hand, the products' side contains

• one atom of copper, $1 \times \text{Cu}$
• two atoms of hydrogen, $2 \times \text{H}$
• two atoms of sulfur, $2 \times \text{S}$
• seven atoms of oxygen, $5 \times \text{O}$

Focus on balancing the sulfur atoms first. Multiply the sulfuric acid by $\textcolor{red}{2}$ to get

${\text{Cu"_ ((s)) + color(red)(2)"H"_ 2"SO"_ (4(aq)) -> "CuSO"_ (4(aq)) + "H"_ 2"O"_ ((l)) + "SO}}_{2 \left(g\right)}$

The reactants' side now contains

• one atom of copper, $1 \times \text{Cu}$
• four atoms of hydrogen, $\textcolor{red}{2} \times 2 \times \text{H}$
• two atom of sulfur, $\textcolor{red}{2} \times 1 \times \text{S}$
• eight atoms of oxygen, $\textcolor{red}{2} \times 4 \times \text{O}$

Now, notice that the products' side is two atoms of hydrogen and one atom of oxygen short. To balance these atoms out, multiply the water molecule by $\textcolor{g r e e n}{2}$

${\text{Cu"_ ((s)) + color(red)(2)"H"_ 2"SO"_ (4(aq)) -> "CuSO"_ (4(aq)) + color(green)(2)"H"_ 2"O"_ ((l)) + "SO}}_{2 \left(g\right)}$

The reactants' side remains unchanged, but the products' side will now contain

• one atom of copper, $1 \times \text{Cu}$
• four atoms of hydrogen, $\textcolor{g r e e n}{2} \times 2 \times \text{H}$
• two atom of sulfur, $2 \times \text{S}$
• eight atoms of oxygen, $\left(\textcolor{g r e e n}{2} + 4 + 2\right) \times \text{O}$

Since you have equal numbers of atoms on both sides of the equation, you can say that the chemical equation is balanced.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{Cu"_ ((s)) + 2"H"_ 2"SO"_ (4(aq)) -> "CuSO"_ (4(aq)) + 2"H"_ 2"O"_ ((l)) + "SO}}_{2 \left(g\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$