# Question ed921

Jun 23, 2017

The reaction produces 2.84 L of ${\text{CO}}_{2}$.

#### Explanation:

$\text{2C"_2"H"_2 + "5O"_2 → "4CO"_2 + "2H"_2"O}$

Step 2. Calculate the moles of ${\text{O}}_{2}$

${\text{Moles of O"_2 = 5.00 color(red)(cancel(color(black)("g O"_2))) × "1 mol O"_2/(32.00 color(red)(cancel(color(black)("g O" _2)))) = "0.1562 mol O}}_{2}$

Step 3. Calculate the moles of ${\text{CO}}_{2}$

${\text{Moles of CO"_2 = 0.1562 color(red)(cancel(color(black)("mol O"_2))) × "4 mol CO"_2/(5 color(red)(cancel(color(black)("mol O"_2)))) = "0.1250 mol O}}_{2}$

Step 4. Use the Ideal Gas Law to calculate the moles of ${\text{CO}}_{2}$

The Ideal Gas Law is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

• $p$ is the pressure
• $V$ is the volume
• $n$ is the number of moles
• $R$ is the gas constant
• $T$ is the temperature

We can rearrange the Ideal Gas Law to get

$V = \frac{n R T}{p}$

STP is 1 bar and 0 °C.

V = (0.1250 color(red)(cancel(color(black)("mol"))) × "0.083 14" color(red)(cancel(color(black)("bar")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "2.84 L"#