Question

12) How many liters of pure oxygen gas, measured at STP, are required for the complete reaction with 4.05 L of C4H10(g), also measured at STP?2 C4H10(g) + 13 O2(g) --> 8 CO2(g) + 10 H2O(g)

Answer 1

Answer:

26.2L of O2 gas are required.

Explanation:

1st) It is necessary to convert the 4.05L to moles. One mole of a gaseous substance occupies a volume of 22.4L, so wit a mathematical rule of three cwe can calculate the number of moles:

$\begin{gathered} 22.4L-1mole \\ 4.05L-x=\frac{4.05L*1mole}{22.4L} \\ x=0.18moles \end{gathered}$

2nd) From the balanced reaction, we know that 2 moles of C4H10 gas react with 13 moles of O2 gas. So, with the 0.18 moles of C4H10 we need to calculate the amount of O2 gas that will react:

$\begin{gathered} 2molesC_4H_{10}-13molesO_2 \\ 0.18molesC_4H_{10}-x=\frac{0.18molesC_4H_{10}*13molesO_2}{2molesC_4H_{10}} \\ x=1.17molesO_2 \end{gathered}$

Now we know that 1.17 moles of O2 gas are required.

3rd) Finally, we have to convert the 1.17 moles to liters, using the 1mole:22.4L conversion:

$\begin{gathered} 1mole-22.4L \\ 1.17moles-x=\frac{1.17moles*22.4L}{1mole} \\ x=26.2L \end{gathered}$

So, 26.2L of O2 gas are required.