Question

6. What volume of CO2 gas at STP, would be present in 80.55 g?

Answer 1

Answer:

41.0 L CO₂

Explanation:

We want to determine the volume of carbon dioxide gas present in a 80.55 gram sample at STP.

To do so, we can convert from grams to moles and then use the STP ratio to determine its volume.

Find the molar mass of CO₂:

$\displaystyle\begin{aligned} \text{MM}_{\text{CO _2 }}&= (12.01 + 2(16.00)) \text{ g/mol} \\ \\ &= 44.01 \text{ g/mol} \end{aligned}$

And recall that for any gas at STP, there are 22.4 L/mol.

Dimensional Analysis:

$\displaystyle 80.55\text{ g CO _2 }\cdot \frac{1 \text{ mol CO _2 }}{44.01 \text{ gCO _2 }} \cdot \frac{22.4 \text{ L CO _2 }}{1 \text{ mol CO _2 }} = 41.0 \text{ L CO _2 }$

In conclusion, there will be about 41.0 L of carbon dioxide gas at STP in a 80.55 gram sample.