First, we need to determine the number of moles of [tex]\( \text{C}_2\text{H}_6 \)[/tex] in 1.56 grams. To do this, we use the molar mass of [tex]\( \text{C}_2\text{H}_6 \)[/tex].
The molar mass of [tex]\( \text{C}_2\text{H}_6 \)[/tex] is calculated as follows:
- Carbon: [tex]\( 12.01 \)[/tex] g/mol
- Hydrogen: [tex]\( 1.01 \)[/tex] g/mol
Thus, the molar mass of [tex]\( \text{C}_2\text{H}_6 \)[/tex] is:
[tex]\[ 2 \times 12.01 + 6 \times 1.01 = 24.02 + 6.06 = 30.08 \, \text{g/mol} \][/tex]
Now, we calculate the moles of [tex]\( \text{C}_2\text{H}_6 \)[/tex]:
[tex]\[ \text{Moles of } \text{C}_2\text{H}_6 = \frac{1.56 \, \text{g}}{30.08 \, \text{g/mol}} \approx 0.0519 \text { moles } \][/tex]
The balanced chemical equation [tex]\( \text{2 C}_2\text{H}_6 + 7 \text{O}_2 \rightarrow 4 \text{CO}_2 + 6 \text{H}_2\text{O} \)[/tex] shows that 2 moles of [tex]\( \text{C}_2\text{H}_6 \)[/tex] produce 4 moles of [tex]\( \text{CO}_2 \)[/tex]. Therefore, 1 mole of [tex]\( \text{C}_2\text{H}_6 \)[/tex] will produce 2 moles of [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[ \text{Moles of } \text{CO}_2 = 0.0519 \, \text{moles of } \text{C}_2\text{H}_6 \times 2 \approx 0.1037 \text { moles } \][/tex]
Next, we calculate the mass of [tex]\( \text{CO}_2 \)[/tex]. The molar mass of [tex]\( \text{CO}_2 \)[/tex] is:
- Carbon: [tex]\( 12.01 \)[/tex] g/mol
- Oxygen: [tex]\( 16.00 \)[/tex] g/mol
Thus, the molar mass of [tex]\( \text{CO}_2 \)[/tex] is:
[tex]\[ 12.01 + 2 \times 16.00 = 12.01 + 32.00 = 44.01 \, \text{g/mol} \][/tex]
Now, we determine the mass of [tex]\( \text{CO}_2 \)[/tex] produced:
[tex]\[ \text{Mass of } \text{CO}_2 = 0.1037 \, \text{moles} \times 44.01 \, \text{g/mol} \approx 4.56 \text { grams } \][/tex]
Hence, when 1.56 grams of [tex]\( \text{C}_2\text{H}_6 \)[/tex] reacts with excess oxygen, approximately 4.56 grams of [tex]\( \text{CO}_2 \)[/tex] are formed.
