Using the balanced equation for the combustion of ethylene [tex]$\left( C_2H_4 \right)$[/tex], answer the following questions.

[tex]\[
C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O
\][/tex]

Be sure each of your answers has the correct number of significant figures.

Part 1 of 2:

How many grams of [tex]$CO_2$[/tex] are formed from [tex]$1.9 \text{ mol}$[/tex] of [tex]$C_2H_4$[/tex]?

[tex]$\square$[/tex] [tex]$g \, CO_2$[/tex] [tex]$\square$[/tex] [tex]$\square \times 10$[/tex]



Answer :

To determine the grams of CO₂ formed from 1.9 moles of C₂H₄, follow these detailed steps:

1. Understanding the Chemical Equation:
- The balanced equation for the combustion of ethylene (C₂H₄) is:
[tex]\[ C_2H_4 + 3 O_2 \rightarrow 2 CO_2 + 2 H_2O \][/tex]
- This equation shows that 1 mole of C₂H₄ produces 2 moles of CO₂.

2. Moles of CO₂ Produced:
- Given 1.9 moles of C₂H₄, we can determine the moles of CO₂ produced using the stoichiometric relationship from the equation. Since 1 mole of C₂H₄ produces 2 moles of CO₂, 1.9 moles of C₂H₄ will produce:
[tex]\[ \text{moles of CO₂} = 1.9 \text{ mol C₂H₄} \times 2 \text{ mol CO₂/mol C₂H₄} = 3.8 \text{ mol CO₂} \][/tex]

3. Molar Mass of CO₂:
- Calculate the molar mass of CO₂:
[tex]\[ \text{molar mass of CO₂} = 12.01 \text{ (C) } + 2 \times 16.00 \text{ (O) } = 44.01 \text{ g/mol} \][/tex]

4. Grams of CO₂ Produced:
- To find the mass of CO₂ in grams, multiply the moles of CO₂ by its molar mass:
[tex]\[ \text{grams of CO₂} = 3.8 \text{ mol CO₂} \times 44.01 \text{ g/mol} = 167.23799999999997 \text{ g} \][/tex]

5. Significant Figures:
- The given data (1.9 mol) has 2 significant figures.
- Thus, the final answer should also be reported with an appropriate number of significant figures, which in this case, is 2 significant figures.

6. Final Answer in Correct Format:
- Rounded to 2 significant figures, the mass of CO₂ produced is:
[tex]\[ 1.7 \times 10^2 \text{ grams of CO₂} \][/tex]

Thus, 167.24 g CO₂ is produced.

Other Questions