# Answer in General Chemistry for mark #279316

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 }_{) }

**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) u2192 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) u2192 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 H_{2}O.

“C_4H_{10} (g) + O_2(g) u2192 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) u2192 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) u2192 4 CO_2(g) + 5 H_2O(l ) ]”

“2 C_4H_{10} (g) + 13 O_2(g) u2192 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) u2192 8 CO_2(g) + 10 H_2O(l ) }”

Now we will find the volume of H_{2}O in Liters formed from combustion of 14.9 L of butane ( C_{4}H_{10})

In the given reaction you can see that 2 volume of C_{4}H_{10} reacts with 13 volume of O_{2} to give 8 volume of CO_{2} and 10 volume of H_{2}O.

So volume of H_{2}O 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”