Question

acetylene (c₂h₂) gas is often used in welding torches because of the very high heat produced when it reacts with oxygen (o₂) gas, producing carbon dioxide gas and water vapor. calculate the moles of water produced by the reaction of 0.500 mol of acetylene. be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.

Explanation:

Step1: Write the balanced chemical equation

The reaction of acetylene (\(C_2H_2\)) with oxygen (\(O_2\)) to form carbon dioxide (\(CO_2\)) and water (\(H_2O\)) is:

$$2C_2H_2 + 5O_2 ightarrow4CO_2 + 2H_2O$$

Step2: Determine the mole ratio

From the balanced equation, the mole ratio of \(C_2H_2\) to \(H_2O\) is \(2:2 = 1:1\).

Step3: Calculate the moles of water

Given \(n(C_2H_2)=0.500\space mol\). Using the mole ratio \(\frac{n(H_2O)}{n(C_2H_2)}=\frac{1}{1}\), we have \(n(H_2O)=n(C_2H_2)\times\frac{1}{1}\).
Substituting the value of \(n(C_2H_2)\): \(n(H_2O)=0.500\space mol\times1 = 0.500\space mol\)

Answer:

\(0.500\space mol\)