Question

How many liters of O₂ at STP is needed to completely burn 20.0 L of butane(C₄H₁₀), with this reaction?

2C₄H₁₀+13O₂ ---> 8CO₂+10H₂O

Answer 1

Solution:

given chemical reaction:

$2C_4H_{10}+13O_2 \to 8CO_2 +10 H_2O$

I) We need to divide given volume of butane in order to find moles of it:

$n=\frac{20}{22.4}=0.8928moles\;of\;butane$

$n_{O_2}=\frac{0.8928 \times 13}{2}=5.799\;moles \;of\;oxygen$

II) We need to multiply moles of oxygen by volume in standard conditions

$V_{O_2}=n \times V _o=5.799 \times 22.4 L$

$V_{O_2}=129.91L$

Answer: We need 129.91 L of oxygen to completely burn 20.0 L of butane.