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

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

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

Part 1 of 2:

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

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

Part 2 of 2:

How many grams of [tex]$H_2O$[/tex] are formed from 0.35 mol of [tex]$C_2H_4$[/tex]?

[tex]$\square \, g \, H_2O$[/tex]



Answer :

GinnyAnswer
Let's solve each part of this question step-by-step using stoichiometry from the balanced equation of the combustion of ethylene \(\left( \text{C}_2\text{H}_4\right)\):

[tex]\[ \text{C}_2\text{H}_4 + 3 \text{O}_2 \rightarrow 2 \text{CO}_2 + 2 \text{H}_2\text{O} \][/tex]

Part 1: Grams of \(\text{CO}_2\) formed from 1.9 mol of \(\text{C}_2\text{H}_4\)

1. Identify the molar ratio between \(\text{C}_2\text{H}_4\) and \(\text{CO}_2\):
According to the balanced equation, 1 mole of \(\text{C}_2\text{H}_4\) produces 2 moles of \(\text{CO}_2\).

2. Calculate the moles of \(\text{CO}_2\) formed:
Since 1.9 moles of \(\text{C}_2\text{H}_4\) are given:
[tex]\[ \text{Moles of \(\text{CO}_2\)} = 1.9 \text{ mol \(\text{C}_2\text{H}_4\)} \times 2 \frac{\text{mol \(\text{CO}_2\)}}{\text{mol \(\text{C}_2\text{H}_4\)}} \][/tex]
[tex]\[ \text{Moles of \(\text{CO}_2\)} = 3.8 \text{ mol} \][/tex]

3. Calculate the grams of \(\text{CO}_2\) formed:
The molar mass of \(\text{CO}_2\) (carbon dioxide) is 44.01 g/mol.
[tex]\[ \text{Mass of \(\text{CO}_2\)} = 3.8 \text{ mol} \times 44.01 \frac{\text{g}}{\text{mol}} \][/tex]
[tex]\[ \text{Mass of \(\text{CO}_2\)} = 167.24 \text{ g} \][/tex]

Thus, the number of grams of \(\text{CO}_2\) formed from 1.9 mol of \(\text{C}_2\text{H}_4\) is:
[tex]\[ 167.24 \text{ g CO}_2 \][/tex]

Part 2: Grams of \(\text{H}_2\text{O}\) formed from 0.35 mol of \(\text{C}_2\text{H}_4\)

1. Identify the molar ratio between \(\text{C}_2\text{H}_4\) and \(\text{H}_2\text{O}\):
According to the balanced equation, 1 mole of \(\text{C}_2\text{H}_4\) produces 2 moles of \(\text{H}_2\text{O}\).

2. Calculate the moles of \(\text{H}_2\text{O}\) formed:
Since 0.35 moles of \(\text{C}_2\text{H}_4\) are given:
[tex]\[ \text{Moles of \(\text{H}_2\text{O}\)} = 0.35 \text{ mol} \times 2 \frac{\text{mol \(\text{H}_2\text{O}\)}}{\text{mol \(\text{C}_2\text{H}_4\)}} \][/tex]
[tex]\[ \text{Moles of \(\text{H}_2\text{O}\)} = 0.7 \text{ mol} \][/tex]

3. Calculate the grams of \(\text{H}_2\text{O}\) formed:
The molar mass of \(\text{H}_2\text{O}\) (water) is 18.02 g/mol.
[tex]\[ \text{Mass of \(\text{H}_2\text{O}\)} = 0.7 \text{ mol} \times 18.02 \frac{\text{g}}{\text{mol}} \][/tex]
[tex]\[ \text{Mass of \(\text{H}_2\text{O}\)} = 12.614 \text{ g} \][/tex]

Thus, the number of grams of \(\text{H}_2\text{O}\) formed from 0.35 mol of \(\text{C}_2\text{H}_4\) is:
[tex]\[ 12.614 \text{ g H}_2\text{O} \][/tex]
Thus in grams:
[tex]\[ 167.24 \text{ g CO}_2, \quad 12.614 \text{ g H}_2\text{O} \][/tex]

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