43.7 L of butane (C4H10) measured at S.T.P is burned in excess oxygen. Write an equation for the reaction and what is the volume of carbon dioxide measured at S.T.P?

1 Answer
Aug 16, 2017

Answer:

#C_4H_10(g) + 13/2O_2(g) rarr4CO_2(g) + 5H_2O(g)#

We get approx. #176*L# of carbon dioxide gas.

Explanation:

The normal rigmarole is to balance the carbons as carbon dioxide, the hydrogens as water, and then balance the oxygens on the reactant side. If you like you can double the entire reaction to give integral coefficients, i.e.

#2C_4H_10(g) + 13O_2(g) rarr8CO_2(g) + 10H_2O(g)#

I usually prefer the FORMER representation in that the stoichiometry is EASIER to rationalize. And of course while we cannot have half an oxygen molecule, CERTAINLY we can have a #16*g# quantity of dioxygen gas.

And given #C_4H_10(g) + 13/2O_2(g) rarr4CO_2(g) + 5H_2O(g)# if we completely combust a #43.7*L# volume of butane, NECESSARILY we gets a #4xx43.7*L# volume of carbon dioxide, i.e. approx. #176*L# #CO_2(g)#.

I have assumed that the molar volume of an Ideal Gas is #22.4*L*mol^-1# at #"STP"#.