Question

Potassium nitrate$\mathbf{\left(}\mathit{K}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{\right)}$is used as a fertilizer for certain crops. It uses produced through the reaction

$\mathbf{4}\mathit{K}\mathit{C}\mathit{l}\mathbf{+}\mathbf{4}\mathit{H}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{\to }\mathbf{4}\mathit{K}\mathit{N}{\mathit{O}}_{\mathbf{3}}\mathbf{+}\mathbf{2}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathit{a}$

Calculate the minimum mass of KCl required toproduce$\mathbf{567}\mathit{g}$$\mathit{K}\mathit{N}{\mathit{O}}_{\mathbf{3}}$. What mass of$\mathit{C}{\mathit{l}}_{\mathbf{2}}$will be generated as well?

Short answer

$418g$of potassium chloride and $199g$of chlorine gas.

Step-by-step solution

Formation of potassium nitrate

Potassium nitrate is formed by reacting potassium chloride, nitric acid, and oxygen gas. The other products formed are chlorine and water.

• The molar mass of KCl is 74.6 g/mol.
• The molar mass ofpotassium nitrateis 101.1 g/mol.
• The molar mass ofis 70.9 g/mol.

Using the unitary method

According to the balanced chemical reaction, the following relations are true:

$2molKCl\to 2molKN{O}_3$and$1molC{l}_2$

$2\times 74.6g\to 2\times 101.1g$and$70.9g$

$?g\to 567g$and y gram

The unknown masses of potassium chloride as reactant are needed, and the byproduct chlorine can be calculated using the direct proportion as stated above.

Calculating the requiredmasses

The direct proportion can be solved for the mass of the reactant as an equation:

$$\begin{gathered} m_{KCl}=\frac{567}{202.2}\times 149.2 \\ =418g \end{gathered}$$

And the direct proportion can be solved for the mass of the byproduct as an equation:

$$\begin{gathered} m_{C{l}_2}=\frac{567}{202.2}\times 70.9 \\ =199g \end{gathered}$$

Therefore, $418g$of potassium chloride and $199g$of chlorine gas.