Could someone please help me?

Question

Using the following balanced equation: 

4K + O₂ → 2K₂O

 How many grams of potassium oxide (K₂O) are formed from 8.1 g of oxygen (O₂) reacting with excess potassium?

Answer 1

4K + O₂ → 2K₂O ... balanced equation

Since potassium (K) is in excess, the amount (MOLES) of O₂ present will dictate how much product (K₂O) can be formed. Work in moles, not grams, and then at the end, convert to grams.

Moles of O₂ present = 8.1 g O₂ x 1 mole O₂ / 32 g = 0.2531 moles O₂ present

Use the mole ratio in the balanced equation to find the MOLES of K₂O formed:

0.2531 moles O₂ x 2 moles K₂O / 1 mole O₂ = 0.5062 moles K₂O

Convert moles K₂O to grams:

0.5062 moles K₂O x 94.17 g K₂O / mole K₂O = 47.67 g K₂O formed

Correcting for significant figures:

48 g K₂O formed (2 sig.figs. based on 8.1 g O₂ used)

Answer 2

Hello, thank you for taking the time to post your question!

First you want to take the 8.1 grams of O2 and convert it over into moles by dividing by the molar mass

8.1 g / 32 g/mol = 0.2531 mol O2

Then you can convert that over into moles of K2O using the coefficients from the equation

0.2531 x 2 mol K2O/1 mol O2 = 0.5062 mol K2O

Finally then you want to convert that over into grams by multiplying by the molar mass

0.5062 mol K2O x 94.20 g/mol = 47.69 g K2O

So it ends up forming 47.69 grams of potassium oxide

I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)