Answer to Question #279316 in General Chemistry for mark

Question

Assuming no change in temperature and pressure, Calculate the volume of O2 (in liters) required for the complete combustion of 14.9 L of butane (C4H10):

                                        C4H10_(g) + O2_(g)       →  CO2_(g) + H2O_{(l )}    

Answer 1

Q279316

Assuming no change in temperature and pressure, Calculate the volume of O2 (in liters) required for the complete combustion of 14.9 L of butane (C4H10):

                                        C4H10_(g) + O2_(g)       →  CO2_(g) + H2O_{(l )}   

Solution :

 

$$C_4H_{10} (g) + O_2(g) → CO_2(g) + H_2O(l )$$

Let us first balance this reaction.

there is 4 C on left side and 1 C on right side. If we write 4 in front of CO ₂, then both side will contain same number of C.

$$C_4H_{10} (g) + O_2(g) → 4 \ CO_2(g) + H_2O(l )$$

There is 10 H on left side and 2 H on right side. For balancing H we will write 5 in front of H₂O.

$$C_4H_{10} (g) + O_2(g) → 4 \ CO_2(g) + 5 \ H_2O(l )$$

Lastly we will balance the O on both the side.

There is 2 O on left side and 13 O on right side.

A proper fractional number must be written in front of O2, so that both side contains same number of O.

If we write 13/2 in front of O2, then both side will contain same number of O which is 13.

 

$$C_4H_{10} (g) +\frac{13}{2} O_2(g) → 4 \ CO_2(g) + 5 \ H_2O(l )$$

In the balanced chemical reaction the coefficient can't be fraction.

multiply whole reaction by 2 for removing the fraction.

$$2 * [ \ C_4H_{10} (g) +\frac{13}{2} O_2(g) → 4 \ CO_2(g) + 5 \ H_2O(l ) \ ]$$

$$2 \ C_4H_{10} (g) \ + \ 13 \ O_2(g) → 8 \ CO_2(g) + 10 \ H_2O(l )$$

Now the reaction is balanced for all the elements

$$\textcolor{darkblue}{2 \ C_4H_{10} (g) \ + \ 13 \ O_2(g) → 8 \ CO_2(g) + 10 \ H_2O(l ) }$$

Now we will find the volume of H₂O in Liters formed from combustion of 14.9 L of butane ( C₄H₁₀)

In the given reaction you can see that 2 volume of C₄H₁₀ reacts with 13 volume of O₂ to give 8 volume of CO₂ and 10 volume of H₂O.

So volume of H₂O formed on combustion of 14.9 L of butane is

$$=14.9 \ \cancel{L \ C_4H_{10}} * \frac{10 \ L \ H_2O }{2 \ \cancel{L \ C_4H_{10}} }$$

$$= 74.5 \ L \ H_2O$$

Hence 74.5 L of H₂O will be formed on complete combustion of 14.9 L of butane.