Questions: A reaction between liquid reactants takes place at 28.0°C in a sealed, evacuated vessel with a measured volume of 10.0 L. Measurements show that the reaction produced 6.0 g of carbon dioxide gas. Calculate the pressure of carbon dioxide gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits.

A reaction between liquid reactants takes place at 28.0°C in a sealed, evacuated vessel with a measured volume of 10.0 L. Measurements show that the reaction produced 6.0 g of carbon dioxide gas.

Calculate the pressure of carbon dioxide gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits.

Transcript text: A reaction between liquid reactants takes place at $28.0^{\circ} \mathrm{C}$ in a sealed, evacuated vessel with a measured volume of 10.0 L. Measurements show that the reaction produced 6.0 g of carbon dioxide gas. Calculate the pressure of carbon dioxide gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits. pressure: $\square$ atm

Solution

Solution Steps

Step 1: Convert Temperature to Kelvin

The temperature given is $28.0^{\circ} \mathrm{C}$. To use the ideal gas law, we need to convert this temperature to Kelvin using the formula: \[ T(K) = T(^{\circ}C) + 273.15 \] \[ T = 28.0 + 273.15 = 301.15 \, \text{K} \]

Step 2: Calculate Moles of Carbon Dioxide

The mass of carbon dioxide ($\text{CO}_2$) is given as 6.0 g. We need to convert this mass to moles using the molar mass of $\text{CO}_2$, which is approximately 44.01 g/mol: \[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{6.0 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.1364 \, \text{mol} \]

Step 3: Use the Ideal Gas Law to Find Pressure

The ideal gas law is given by: \[ PV = nRT \] where:

$P$ is the pressure in atm,
$V$ is the volume in liters,
$n$ is the number of moles,
$R$ is the ideal gas constant, $0.0821 \, \text{L atm/mol K}$,
$T$ is the temperature in Kelvin.

Rearranging the formula to solve for pressure $P$: \[ P = \frac{nRT}{V} \] Substituting the known values: \[ P = \frac{0.1364 \, \text{mol} \times 0.0821 \, \text{L atm/mol K} \times 301.15 \, \text{K}}{10.0 \, \text{L}} \] \[ P \approx \frac{3.3685}{10.0} \approx 0.3369 \, \text{atm} \]