Question

when ethane (c₂h₆) burns, it produces carbon dioxide and water:
2c₂h₆ (g) + 7o₂ (g) —→ 4co₂ (g) + 6h₂o (l)
how many liters of carbon dioxide will be produced when 89.5 l of ethane are burned?
(one mole of any gas occupies 22.4 l under certain conditions of temperature and pressure. assume those conditions for this question.)
○ 179 l
○ 313 l
○ 538 l
○ 358 l

Explanation:

Step1: Calculate moles of ethane

Given volume of ethane \(V = 89.5\space L\). Using the formula \(n=\frac{V}{V_m}\) (where \(V_m = 22.4\space L/mol\)), the number of moles of \(C_2H_6\), \(n_{C_2H_6}=\frac{89.5}{22.4}\space mol\)

Step2: Use mole - ratio from balanced equation

From the balanced equation \(2C_2H_6(g)+7O_2(g)\to4CO_2(g) + 6H_2O(l)\), the mole - ratio of \(C_2H_6\) to \(CO_2\) is \(\frac{n_{C_2H_6}}{n_{CO_2}}=\frac{2}{4}=\frac{1}{2}\). So \(n_{CO_2}=2n_{C_2H_6}\)

Step3: Calculate volume of \(CO_2\)

Since \(V = n\times V_m\), substituting \(n_{CO_2}\) into the volume formula. \(V_{CO_2}=n_{CO_2}\times22.4\). Substituting \(n_{C_2H_6}=\frac{89.5}{22.4}\) into \(n_{CO_2}\), we get \(n_{CO_2}=2\times\frac{89.5}{22.4}\). Then \(V_{CO_2}=(2\times\frac{89.5}{22.4})\times22.4 = 2\times89.5=179\space L\)

Answer:

A. \(179\space L\)