Answer: 24 moles of carbon dioxide are produced.
Explanation: Reaction for combustion of butane is given by the reaction:
\(2C_4H_{10}+13O_2\rightarrow 10H_2O+8CO_2\)
Here, the butane is present in lower amount, hence it is considered as limiting reagent because it limits the formation of the product.
By stoichiometry,
2 moles of butane produces 8 moles of carbon dioxide, so
6 moles of butane will produce = \(\frac{8}{2}\times 6=24\)
Hence, 24 moles of carbon dioxide is produced from the combustion of 6 moles of butane.