Chapter 20, Problem 2RQ

To determine

The number of chromium atoms and oxygen atoms on the right side of the given balanced chemical equation:

$4\text{Cr}\left(s\right)+3{\text{O}}_2\left(g\right)\to 2{\text{Cr}}_2{\text{O}}_3\left(g\right)$

Answer to Problem 2RQ

In the given chemical equation, there are four chromium atoms and six oxygen atoms present on the right side.

Explanation of Solution

The given chemical equation is as shown below.

$4\text{Cr}\left(s\right)+3{\text{O}}_2\left(g\right)\to 2{\text{Cr}}_2{\text{O}}_3\left(g\right)$

The exact number of reactant molecules participating and the exact number of product molecules generating in the chemical equation can be depicted by using the numbers associated with the molecular formula. These numbers are called coefficients of the species involved in the chemical equation.

The right side is the product side. The product side has two ${\text{Cr}}_2{\text{O}}_3$ molecules.

So, the number of chromium atoms present on the right side is $2\times 2\text{ Cr}$ atoms thus total $4\text{ Cr}$ atoms.

Similarly, the number of oxygen atoms present on the right side is $2\times 3\text{ O}$ atoms thus total $6\text{ O}$ atoms.

Conclusion:

Therefore, there are four chromium atoms and six oxygen atoms present on the right side of the given balanced equation.