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 H2O.
“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 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”
