1) The number of moles of carbon dioxide produced by the combustion of 1.1 mol of butane, C4H10, in presence of excess oxygen is:
2C4H10 + 13O2 → 8CO2 + 10H2O
2) How many moles of KCl are produced from 1 mol of Cl2 and excess K in the following reaction?
2K + Cl2 → 2KCl
PLEASE HELP ASAP THESE 2 QUESTIONS?
2020-07-14 12:43 am
回答 (1)
2020-07-14 1:27 am
1)
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Mole ratio C₄H₁₀ : CO₂ = 2 : 8
Moles of CO₂ produced = (1.1 mol) × (8/2) = 4.4 mol
OR:
(1.1 mol C₄H₁₀) × (8 mol CO₂ / 2 mol C₄H₁₀)
= 4.4 mol CO₂
====
2)
2K + Cl₂ → 2KCl
Mole ratio Cl₂ : KCl = 1 : 2
Moles of KCl produced = (1 mol) × 2 = 2 mol
OR:
(1 mol Cl₂) × (2 mol KCl / 1 mol Cl₂)
= 2 mol KCl
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O
Mole ratio C₄H₁₀ : CO₂ = 2 : 8
Moles of CO₂ produced = (1.1 mol) × (8/2) = 4.4 mol
OR:
(1.1 mol C₄H₁₀) × (8 mol CO₂ / 2 mol C₄H₁₀)
= 4.4 mol CO₂
====
2)
2K + Cl₂ → 2KCl
Mole ratio Cl₂ : KCl = 1 : 2
Moles of KCl produced = (1 mol) × 2 = 2 mol
OR:
(1 mol Cl₂) × (2 mol KCl / 1 mol Cl₂)
= 2 mol KCl
收錄日期: 2021-04-18 18:33:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200713164339AATnoic