The moles of oxygen formed when 58.6 g of KNO3 decomposes is [tex]\boxed{{\text{0}}{\text{.7245moles}}}[/tex]
Further Explanation:
Stoichiometry of a reaction is used to determine the amount of species present in the reaction by the relationship between the reactants and products. It can be used to determine the moles of a chemical species when the moles of other chemical species present in the reaction is given.
Consider the general reaction,
[tex]{\text{A}}+2{\text{B}}\to3{\text{C}}[/tex]
Here,
A and B are reactants.
C is the product.
One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.
The given reaction is,
[tex]{\text{4KN}}{{\text{O}}_3}\left(s\right)\to2{{\text{K}}_2}{\text{O}}\left(s\right)+2{{\text{N}}_2}\left(g\right)+5{{\text{O}}_2}\left(g\right)[/tex]
On reactant side,
Number of potassium atoms is 4.
Number of nitrogen atom is 4.
Number of oxygen atoms is 12.
On the product side,
Number of potassium atoms is 4.
Number of nitrogen atom is 4.
Number of oxygen atoms is 12.
The number of atoms of all the species in both the reactant and the product side is the same. So above reaction is balanced. The stoichiometry of the balanced reaction indicates that 4 moles of [tex]{\text{KN}}{{\text{O}}_3}[/tex] decompose to give 2 moles of [tex]{{\text{K}}_2}{\text{O}}[/tex] to form 2 moles of [tex]{{\text{N}}_2}[/tex] and 5 moles of [tex]{{\text{O}}_2}[/tex].
The formula to calculate the number of moles of [tex]{\mathbf{KN}}{{\mathbf{O}}_{\mathbf{3}}}[/tex] is as follows:
[tex]{\text{Moles of KN}}{{\text{O}}_3}=\frac{{{\text{Given mass of KN}}{{\text{O}}_3}}}{{{\text{Molar mass of KN}}{{\text{O}}_3}}}[/tex] …… (1)
The given mass of is 58.6 g.
The molar mass of is 101.11 g/mol.
Substitute these values in equation (1)
[tex]\begin{gathered}{\text{Moles of KN}}{{\text{O}}_3}{\mathbf{=}}\left({58.6\;{\text{g}}}\right)\left({\frac{{{\text{1}}\;{\text{mol}}}}{{{\text{101}}{\text{.11}}\;{\text{g}}}}}\right)\\=0.5796\;{\text{mol}}\\\end{gathered}[/tex]
According to the stoichiometry, 4 moles of [tex]{\text{KN}}{{\text{O}}_3}[/tex] decompose to give 2 moles of [tex]{{\text{K}}_2}{\text{O}}[/tex] , 2 moles of {{\text{N}}_2} and 5 moles of [tex]{{\text{O}}_2}[/tex].
So the number of moles of {{\mathbf{O}}_{\mathbf{2}}} formed by 0.5796 moles of [tex]{\mathbf{KN}}{{\mathbf{O}}_{\mathbf{3}}}[/tex] is calculated as follows:
[tex]\begin{gathered}{\text{Moles of }}{{\text{O}}_2}{\mathbf{=}}\left({\frac{{5\;{\text{mol}}}}{{4\;{\text{mol}}}}}\right)\times\left({0.5796\;{\text{mol}}}\right)\\=0.7245\;{\text{mol}}\\\end{gathered}[/tex]
Hence, the moles of [tex]{{\mathbf{O}}_{\mathbf{2}}}[/tex] produced is 0.7245 moles.
Learn more:
1. Bond energy of H-H bond in the given reaction: https://brainly.com/question/7213980
2. What coefficients are required to balance equation: https://brainly.com/question/1971314
Answer details:
Grade: High School
Subject: Chemistry
Chapter: Mole concept
Keywords: stoichiometry, KNO3, K2O, O2, N2, moles, A, B, C, molar mass, reactants, products, 0.5796 moles, 0.7245 moles, potassium, oxygen and nitrogen.
