StudyGPT Just Leveled Up – Say Hello to Knowee! 🚀
Knowee
Questions
Features
Study Tools

2C4H10+1302 → 8CO2 + 10H2Oassuming oxygen is the limiting reagent, how much butane would remain if 20g grams butane burns in 60 g of oxygen

Question

🧐 Not the exact question you are looking for?Go ask a question

Solution

First, we need to determine the molar mass of the reactants and products in the chemical equation.

The molar mass of butane (C4H10) is approximately 58.12 g/mol. The molar mass of oxygen (O2) is approximately 32 g/mol.

Next, we need to convert the given mass of the reactants into moles.

For butane: 20 g ÷ 58.12 g/mol = 0.344 moles For oxygen: 60 g ÷ 32 g/mol = 1.875 moles

The stoichiometry of the reaction tells us that 2 moles of butane react with 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water.

Therefore, the mole ratio of butane to oxygen is 2:13.

To find out how much butane would react with the given amount of oxygen, we can set up a proportion:

2/13 = x/1.875

Solving for x gives us approximately 0.288 moles of butane.

Since we started with 0.344 moles of butane, the amount of butane that would remain after the reaction is:

0.344 moles - 0.288 moles = 0.056 moles

Finally, we convert this amount back into grams using the molar mass of butane:

0.056 moles x 58.12 g/mol = approximately 3.25 g

So, approximately 3.25 g of butane would remain if 20 g of butane burns in 60 g of oxygen, assuming oxygen is the limiting reagent.

This problem has been solved

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.