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.
