Which of the following processes could theoretically produce the maximum mass of oxygen per gram of metal salt reactant?

Question

Quizplus · Accepted Answer

A balanced chemical equation can be used to determine the theoretical yield of a reaction, which is the maximum amount of product that can be produced from a given amount of reactant. In this case, we are looking for the reaction that produces the maximum mass of oxygen per gram of metal salt reactant.

To calculate the theoretical yield of oxygen in each reaction, we need to determine the moles of metal salt reactant and the moles of oxygen produced. We can then divide the mass of oxygen produced by the mass of metal salt reactant to find the mass of oxygen per gram of reactant.

Using this method, we can determine that reaction A produces the most oxygen per gram of reactant:

A) 2 KClO3 -> 2 KCl + 3 O2
1 mol KClO3 produces 3 mol O2
Molar mass of KClO3 = 122.55 g/mol
Molar mass of O2 = 32 g/mol
Theoretical yield of O2 from 1 g KClO3:
(3 mol O2/1 mol KClO3) x (32 g O2/1 mol O2) = 96 g O2/mol KClO3
Mass of O2 per gram of KClO3:
96 g O2/122.55 g KClO3 = 0.783 g O2/g KClO3

For the other reactions:
B) 4 KClO2 + 2 CO2 -> 2 K2CO3 + 3 O2
1 mol KClO2 produces 3/4 mol O2
Molar mass of KClO2 = 122.55 g/mol
Theoretical yield of O2 from 1 g KClO2:
(3/4 mol O2/1 mol KClO2) x (32 g O2/1 mol O2) = 24 g O2/mol KClO2
Mass of O2 per gram of KClO2:
24 g O2/122.55 g KClO2 = 0.196 g O2/g KClO2

C) 2 NaClO2 + 2 HCl + Na2SO4 -> 2 HClO + Na2SO4 + H2O
No oxygen is produced in this reaction, so it is not relevant to the question.

D) 2 HgO -> 2 Hg + O2
1 mol HgO produces 1 mol O2
Molar mass of HgO = 216.59 g/mol
Molar mass of O2 = 32 g/mol
Theoretical yield of O2 from 1 g HgO:
(1 mol O2/1 mol HgO) x (32 g O2/1 mol O2) = 32 g O2/mol HgO
Mass of O2 per gram of HgO:
32 g O2/216.59 g HgO = 0.148 g O2/g HgO

E) NaBrO2 -> NaBr + O2
1 mol NaBrO2 produces 1 mol O2
Molar mass of NaBrO2 = 150.89 g/mol
Molar mass of O2 = 32 g/mol
Theoretical yield of O2 from 1 g NaBrO2:
(1 mol O2/1 mol NaBrO2) x (32 g O2/1 mol O2) = 32 g O2/mol NaBrO2
Mass of O2 per gram of NaBrO2:
32 g O2/150.89 g NaBrO2 = 0.212 g O2/g NaBrO2

Therefore, the best choice for producing the maximum mass of oxygen per gram of metal salt reactant is A, with a theoretical yield of 0.783 g O2 per gram of KClO3.

Which of the Following Processes Could Theoretically Produce the Maximum