Answer to Question #145051 in General Chemistry for hunter

Question

2. In a combustion reaction, butane (C4H10) is burned in a lighter at a rate of 1.50 mol/min.

a) What is the rate at which O2(g) is consumed and CO2(g) is produced?

b) How long does 15.0g of butane take to burn?

Answer 1

2С₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

a) Rate(O₂) $= \frac{13}{2}Rate(С_4H_{10}) = \frac{13}{2}1.50 = 9.75 \;mol/min$

Rate(CO2) $= \frac{8}{2}Rate(С_4H_{10}) = \frac{8}{2}1.50 = 6.0 \;mol/min$

b) m = 15.0 g

M(butane) = 58.12 g/mol

n(butane) $= \frac{15.0}{58.12} = 0.258 \;mol$

Time $= \frac{n}{Rate} = \frac{0.258}{1.50} = 0.172\;min = 10.3 \;sec$