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) → K2MnO4(s) + MnO2(s) + O2(g)
How many grams of KMnO4 would you need to produce 0.27 moles of O2, assuming 100% conversion?

Answer 1

0.54 moles of KMnO4, with a molar mass of 158.04 g/mol, are required to produce 0.27 moles of O2, resulting in a needed mass of 85.344 g of KMnO4 assuming 100% conversion.

Step-by-step explanation:

To calculate the mass of KMnO4 needed to produce 0.27 moles of O2, we first examine the balanced chemical equation:

2KMnO4(s) → K2MnO4(s) + MnO2(s) + O2(g)

According to the stoichiometry of the reaction, 2 moles of KMnO4 produce 1 mole of O2. Therefore, to produce 0.27 moles of O2, we need half that amount of KMnO4 in moles, which is 0.27 moles × (2 moles KMnO4 / 1 mole O2), giving us 0.54 moles of KMnO4.

The molar mass of KMnO4 is 158.04 g/mol. To find the mass required, we multiply the number of moles by the molar mass:

Mass of KMnO4 = 0.54 moles × 158.04 g/mol = 85.344 g of KMnO4

Answer 2

You need 85.32 grams of KMnO₄ to produce 0.27 moles of O2, assuming 100% conversion.

Step-by-step explanation:

The balanced chemical reaction is:

2 KMnO₄ (s) → K₂MnO₄ (s) + MnO₂ (s) + O₂ (g)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles participate in the reaction:

• KMnO₄: 2 moles
• K₂MnO₄: 1 mole
• MnO₂: 1 mole
• O₂: 1 mole

Then you can apply the following rule of three: if by stoichiometry 1 mole of O₂ is produced by 2 moles of KMnO₄, 0.27 moles of O₂ are produced by how many moles of KMnO₄?

`moles of KMnO_(4) =(0.27 moles of O_(2) *2moles of KMnO_(4) )/(1mole of O_(2) )`

moles of KMnO₄= 0.54

The molar mass of KMnO₄ is 158
`(g)/(mol)`.

Then the amount of mass present in 0.54 moles of the compound can be calculated by:

0.54 moles* 158.034
`(g)/(mol)`= 85.32 grams

You need 85.32 grams of KMnO₄ to produce 0.27 moles of O2, assuming 100% conversion.