Pressurized metal gas cylinders are generally used to store commonly used gases in the laboratory. At times, it can be easier to chemically prepare occasionally used gases. For example, oxygen gas can be prepared by heating KMnO₄(s) according to the following chemical reaction:
2KMnO₄(s) $\rarr$ K₂MnO₄(s) + MnO₂(s) + O₂(g)
How many grams of KMnO₄ would you need to produce 0.27 $\underline{\text{moles}}$ of O₂, assuming 100% conversion? The molar mass of KMnO₄ is 158.034 g/mol.

Question

Pressurized metal gas cylinders are generally used to store commonly used gases in the laboratory. At times, it can be easier to chemically prepare occasionally used gases. For example, oxygen gas can be prepared by heating KMnO₄(s) according to the following chemical reaction:
2KMnO₄(s)

\rarr

K₂MnO₄(s) + MnO₂(s) + O₂(g)
How many grams of KMnO₄ would you need to produce 0.27

\underline{\text{moles}}

of O₂, assuming 100% conversion? The molar mass of KMnO₄ is 158.034 g/mol.

Quizplus · Accepted Answer

The chemical formula and reaction seem to be incorrectly formatted, but assuming the reaction produces oxygen gas (O2) and the molar mass of KMnO4 (potassium permanganate, which seems to be the intended compound) is given as 158.034 g/mol, we can calculate the amount needed for 0.27 moles of O2. According to the balanced equation provided (assuming it's for the decomposition of KMnO4), 2 moles of KMnO4 produce 1 mole of O2. Therefore, to produce 0.27 moles of O2, we need 0.54 moles of KMnO4. The mass of KMnO4 required is 0.54 moles * 158.034 g/mol = 85.29836 g. Since none of the provided options match this calculation, the correct answer is E) None of the above.

Pressurized Metal Gas Cylinders Are Generally Used to Store Commonly →\rarr→

Pressurized Metal Gas Cylinders Are Generally Used to Store Commonly $\rarr$