Respuesta :
Answer: The total volume of the gaseous products is 1044.29 L
Explanation:
We are given:
Volume of butane = 116 L
At STP:
22.4 L of volume is occupied by 1 mole of a gas
So, 116 L of volume will be occupied by = [tex]\frac{1}{22.4}\times 116=5.18mol[/tex] of butane
The chemical equation for the combustion of butane follows:
[tex]2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)[/tex]
- For carbon dioxide:
By Stoichiometry of the reaction:
2 moles of butane produces 8 moles of carbon dioxide
So, 5.18 moles of butane will produce = [tex]\frac{8}{2}\times 5.18=20.72mol[/tex] of carbon dioxide
Volume of carbon dioxide at STP = (20.72 × 22.4) = 464.13 L
- For water vapor:
By Stoichiometry of the reaction:
2 moles of butane produces 10 moles of water vapor
So, 5.18 moles of butane will produce = [tex]\frac{10}{2}\times 5.18=25.9mol[/tex] of water vapor
Volume of water vapor at STP = (25.9 × 22.4) = 580.16 L
Total volume of the gaseous products = [464.13 + 580.16] = 1044.29 L
Hence, the total volume of the gaseous products is 1044.29 L
