A sample of 6.50 g of liquid 1‑propanol, $ \text{C}_3\text{H}_8\text{O} $, is combusted with 37.1 g of oxygen gas. Carbon dioxide and water are the products.

1. Write the balanced chemical equation for the reaction. (Physical states are optional.)
2. Identify the limiting reactant.

The balanced chemical formula for the combustion of 1-propanol is: C₃H₈O + 6O₂ → 3CO₂ + 4H₂O. The molar mass calculations show that there are fewer moles of oxygen than needed based on the stoichiometry, hence oxygen is the limiting reactant.

Explanation:

The balanced chemical formula for the combustion of 1-propanol is: C₃H₈O + 6O₂ → 3CO₂ + 4H₂O. This indicates 1 mole of 1-propanol reacts with 6 moles of oxygen to yield 3 moles of carbon dioxide and 4 moles of water. Using the stoichiometry of the reaction, we can determine the limiting reactant. The molar mass of 1‑propanol is about 60.1 g/mol and that of oxygen is 32 g/mol. Thus, we have 6.50/60.1 ≈ 0.108 moles of 1-propanol and 37.1/32 ≈ 1.16 moles of oxygen. Because we need 6 moles of oxygen for every mole of 1-propanol, and we have less than that (1.16 / 0.108 ≈ 10.7), oxygen is the limiting reactant in this reaction.

Learn more about Limiting Reactant here:

https://brainly.com/question/33417913

#SPJ4

A sample of 6.50 g of liquid 1‑propanol, \( \text{C}_3\text{H}_8\text{O} \), is combusted with 37.1 g of oxygen gas. Carbon dioxide and water are the products.1. Write the balanced chemical equation for the reaction. (Physical states are optional.)2. Identify the limiting reactant.

Answer :

Final answer:

Explanation:

Learn more about Limiting Reactant here:

Other Questions

A sample of 6.50 g of liquid 1‑propanol, \( \text{C}_3\text{H}_8\text{O} \), is combusted with 37.1 g of oxygen gas. Carbon dioxide and water are the products.

1. Write the balanced chemical equation for the reaction. (Physical states are optional.)
2. Identify the limiting reactant.