Question

consider the combustion of butane (c₄h₁₀):
2c₄h₁₀(g) + 13o₂(g) → 8co₂(g) + 10h₂o(l)
in a particular reaction, 7.05 moles of c₄h₁₀ are reacted with an excess of o₂. calculate the number of moles of co₂ formed. round your answer to 3 significant digits.
mol co₂

Explanation:

Step1: Determine mole ratio

From the balanced equation \(2C_{4}H_{10}(g)+13O_{2}(g)\to8CO_{2}(g) + 10H_{2}O(l)\), the mole ratio of \(C_{4}H_{10}\) to \(CO_{2}\) is \(2:8=\frac{2}{8}=\frac{1}{4}\).

Step2: Calculate moles of \(CO_{2}\)

Let \(n_{CO_{2}}\) be the moles of \(CO_{2}\). We know \(n_{C_{4}H_{10}} = 7.05\) mol. Using the mole - ratio \(\frac{n_{C_{4}H_{10}}}{n_{CO_{2}}}=\frac{2}{8}\), we can solve for \(n_{CO_{2}}\).

$$n_{CO_{2}}=n_{C_{4}H_{10}}\times\frac{8}{2}$$

Substitute \(n_{C_{4}H_{10}} = 7.05\) mol into the equation:

$$n_{CO_{2}}=7.05\times\frac{8}{2}=7.05\times4 = 28.2$$

Answer:

\(28.2\)