Identifying Limiting Reactant and Calculating Mass of Carbon Dioxide in a Chemical Reaction

Introduction

In chemistry, the limiting reactant is the reactant that determines the amount of product that can be formed in a chemical reaction. In this scenario, we have an 8.6 g sample of methane (CH4) and a 15.6 g sample of oxygen (O2) reacting to form carbon dioxide (CO2). By identifying the limiting reactant, we can calculate the mass of carbon dioxide that could be formed.

Limiting Reactant Identification

Given: 22.8 g of CO2 from 8.6 g CH4 and 18.7 g of CO2 from 15.6 g O2

The calculations show that 15.6 g of oxygen is the limiting reactant, as it only produces 10.7 g of CO2.

Calculating Mass of Carbon Dioxide from CH4 and O2

Given: 23.6 g CO2 from 8.6 g CH4 and 10.7 g CO2 from 15.6 g O2

Once again, the calculations confirm that 15.6 g of oxygen is the limiting reactant, producing only 10.7 g of CO2.

Final Calculation and Explanation

Answer:

23.6 g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from 15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of CO2.


Explanation:

1) Balanced chemical equation: CH4 + 2O2 → CO2 + 2H2O

2) Mole ratios: 1 mol CH4 : 2 mol O2 : 1 mol CO2 : 2 mol H2O

3) Molar masses: CH4 – 16.04 g/mol, O2 – 32.0 g/mol, CO2 – 44.01 g/mol

4) Convert reactant masses to moles: CH4 – 0.5362 moles, O2 – 0.4875 moles

5) Mass of CO2 from CH4: 23.60 g

6) Mass of CO2 from O2: 10.7 g

7) O2 is the limiting reactant due to mole ratio constraints

8) Confirmation of CO2 mass from O2: 10.7 g

What is the limiting reactant in the given chemical reaction of methane and oxygen? The limiting reactant in the chemical reaction is oxygen (O2) as it gets used up and only produces 10.7 g of carbon dioxide (CO2).
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