Question

what is the mass in grams of co₂ that can be produced from the combustion of 4.87 moles of butane according to this equation: 2 c₄h₁₀ (g) + 13 o₂ (g) → 8 co₂ (g) + 10 h₂o (g)

Explanation:

Step1: Determine the mole ratio

From the balanced equation \(2\ce{C4H10}(g)+13\ce{O2}(g)\to8\ce{CO2}(g) + 10\ce{H2O}(g)\), the mole ratio of \(\ce{C4H10}\) to \(\ce{CO2}\) is \(2:8\) or \(1:4\).

Step2: Calculate the moles of \(\ce{CO2}\)

If there are \(n(\ce{C4H10})=4.87\space mol\), then \(n(\ce{CO2})=4.87\times4 = 19.48\space mol\)

Step3: Calculate the mass of \(\ce{CO2}\)

The molar mass of \(\ce{CO2}\), \(M(\ce{CO2})=(12 + 2\times16)\space g/mol=44\space g/mol\)
Using the formula \(m = n\times M\), where \(n = 19.48\space mol\) and \(M = 44\space g/mol\)
\(m(\ce{CO2})=19.48\times44\space g\)

$$m(\ce{CO2})=(20 - 0.52)\times44=(20\times44-0.52\times44)=880 - 22.88= 857.12\space g$$

Answer:

\(857.12\space g\)