prob.: Consider the combustion of butane (C4H10):
2C4H10(g) + 13O2(g) ==> 8CO2(g) + 10 H2O(L)
IN A PARTICULAR REACTION, 5.0 MOLES OF C4H10 ARE REACTED
WITHAN EXCESS OF O2. CALCULATE THE NUMBER OF MOLES OF CO2
FORMED.

Respuesta :

Answer: The moles of carbon dioxide formed in the reaction is 20 moles.

Explanation:

We are given:

Number of moles of butane = 5.0 moles

The chemical reaction for the combustion of butane follows the equation:

[tex]2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(l)[/tex]

As, oxygen is present in excess. So, it is considered as an excess reagent.

Thus, butane is considered as a limiting reagent because it limits the formation of products.

By stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide.

So, 5 moles of butane will produce = [tex]\frac{8}{2}\times 5=20mol[/tex] of carbon dioxide.

Hence, the moles of carbon dioxide formed in the reaction is 20 moles.