Answer:
Step-by-step explanation:
The balanced chemical equation for the combustion of ethane (\(C_2H_6\)) in oxygen (\(O_2\)) is as follows:
\[2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O\]
This equation tells us that when 2 moles of ethane (\(C_2H_6\)) react with 7 moles of oxygen (\(O_2\)), it produces 4 moles of carbon dioxide (\(CO_2\)).
To find out how many moles of \(CO_2\) are produced when 5.45 moles of ethane are burned, you can set up a proportion using the mole ratio from the balanced equation:
\[\frac{5.45 \text{ moles of } C_2H_6}{2 \text{ moles of } C_2H_6} = \frac{x \text{ moles of } CO_2}{4 \text{ moles of } CO_2}\]
Now, solve for \(x\ ( \text{moles of } CO_2)\):
\[x = \frac{5.45 \times 4}{2} = 10.9 \text{ moles of } CO_2\]
So, when 5.45 moles of ethane are burned in excess oxygen, 10.9 moles of carbon dioxide are produced.