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]

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]