# How many liters of carbon dioxide can be produced if 37.8 grams of carbon disulfide react with excess oxygen gas at 28.85 degrees Celsius and 1.02 atmospheres?

Jul 6, 2016

$12.1 \setminus L$

#### Explanation:

Carbon disulfide reacts with oxygen to form carbon dioxide and sulfur dioxide.

$C {S}_{2} + {O}_{2} \to C {O}_{2} + S {O}_{2}$

Start by finding the number of moles of the carbon dioxide produced.

37.8 \ "g" \ CS_2 xx ( 1\ "mol."\ CS_2)/(76.1407 \ "g" \ CS_2)xx ("1 mol."\ CO_2)/( 1 \ "mol."\ CS_2)

37.8 \cancel ( "g" \ CS_2) xx ( color(red) cancel( 1\ "mol."\ CS_2))/(76.1407 \cancel( "g" \ CS_2))xx ("1 mol."\ CO_2)/(color (red) cancel( 1\ "mol."\ CS_2))

$0.496 \setminus m o l . \setminus C {O}_{2}$

Now use the ideal gas equation to figure out the the volume of the $C {O}_{2}$ produced.

$P \cdot V = n \cdot R \cdot T$

Where

$P \text{ }$ is the pressure expressed in atm.

$V \text{ }$ is the volume expressed in L.

$n \text{ }$ is the number of moles.

$R \text{ }$ is the universal gas constant, it has a value of $0.0821 \setminus L \cdot a t m . m o {l}^{-} 1 \cdot {K}^{-} 1$

$T \text{ }$is the Kelvin temperature.

Now rearrange the ideal gas equation and solve for V.

$V = \frac{n \cdot R \cdot T}{P}$

$V = \frac{0.496 \setminus m o l \times 0.0821 \setminus L \cdot a t m \cdot m o {l}^{-} 1 \cdot {K}^{-} 1 \times 302.00 \setminus K}{1.02 \setminus a t m}$

$V = \frac{0.496 \setminus \cancel{m o l} \times 0.0821 \setminus L \cdot \cancel{a t m} \cdot \cancel{m o {l}^{-} 1} \cdot \cancel{{K}^{-} 1} \times 302.00 \setminus \cancel{K}}{1.02 \setminus \cancel{a t m}}$

$V = 12.1 \setminus L$