Question

acetylene ($c_{2}h_{2}$) gas is often used in welding torches because of the very high heat produced when it reacts with oxygen ($o_{2}$) gas, producing carbon - dioxide gas and water vapor. calculate the moles of carbon dioxide produced by the reaction of 1.70 mol of oxygen. be sure your answer has a unit symbol, if necessary, and round it to 3 significant digits.

Explanation:

Step1: Write the balanced chemical equation

The reaction of acetylene ($C_2H_2$) with oxygen ($O_2$) is $2C_2H_2 + 5O_2
ightarrow4CO_2+2H_2O$.

Step2: Determine the mole - ratio

From the balanced equation, the mole - ratio of $O_2$ to $CO_2$ is 5:4.

Step3: Calculate the moles of $CO_2$

Let $n_{CO_2}$ be the moles of $CO_2$ and $n_{O_2}$ be the moles of $O_2$. We know that $\frac{n_{CO_2}}{n_{O_2}}=\frac{4}{5}$. Given $n_{O_2} = 1.70$ mol. Then $n_{CO_2}=\frac{4}{5}\times n_{O_2}=\frac{4}{5}\times1.70$ mol.
$n_{CO_2}=1.36$ mol

Answer:

1.36 mol