Question

consider the combustion reaction for acetylene.

$$2\\ce{c2h2(l)} + 5\\ce{o2(g)} \ ightarrow 4\\ce{co2(g)} + 2\\ce{h2o(g)}$$

if the acetylene tank contains 37.0 mol of \\(\ce{c2h2}\\) and the oxygen tank contains 81.0 mol of \\(\ce{o2}\\), what is the limiting reactant for this reaction?

\\(\bigcirc\\) \\(\ce{c2h2}\\)
\\(\bigcirc\\) \\(\ce{o2}\\)
\\(\bigcirc\\) \\(\ce{co2}\\)
\\(\bigcirc\\) \\(\ce{h2o}\\)

Explanation:

Step1: Find moles of O₂ required for C₂H₂

From the reaction \(2\text{C}_2\text{H}_2 + 5\text{O}_2
ightarrow 4\text{CO}_2 + 2\text{H}_2\text{O}\), the mole ratio of \(\text{C}_2\text{H}_2\) to \(\text{O}_2\) is \(2:5\). For \(37.0\) mol of \(\text{C}_2\text{H}_2\), moles of \(\text{O}_2\) required: \(\frac{5}{2} \times 37.0 = 92.5\) mol.

Step2: Compare required and available O₂

Available \(\text{O}_2\) is \(81.0\) mol, which is less than \(92.5\) mol. So \(\text{O}_2\) is limiting. Also, \(\text{CO}_2\) and \(\text{H}_2\text{O}\) are products, not reactants, so they can't be limiting.

Answer:

B. \(\text{O}_2\) (assuming the options are labeled with \(\text{O}_2\) as the correct choice for limiting reactant)