4.4 g of `CO_(2)` contains how many litres of `CO_(2)` at STP ?

Options

A. 2.4 litre
B. 2.24 litre
C. 44 litre
D. 22.4 litre

Explanation

To find out how many liters of \( CO_2 \) are contained in 4.4 g at STP (Standard Temperature and Pressure), we can follow these steps:

### Step 1: Calculate the number of moles of \( CO_2 \)
To find the number of moles, we use the formula:
\[ \text{Number of moles} (n) = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of \( CO_2 \) (Carbon Dioxide) is calculated as follows:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol (and there are 2 oxygen atoms in \( CO_2 \))

So, the molar mass of \( CO_2 \):
\[ \text{Molar mass of } CO_2 = 12 + (2 \times 16) = 12 + 32 = 44 \text{ g/mol} \]

Now, substituting the values into the formula:
\[ n = \frac{4.4 \text{ g}}{44 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 2: Use the Ideal Gas Law to find the volume
The Ideal Gas Law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure (1 atm or 760 mmHg at STP)
- \( V \) = volume (in liters)
- \( n \) = number of moles (calculated in Step 1)
- \( R \) = universal gas constant \( 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \)
- \( T \) = temperature (273 K at STP)

Rearranging the Ideal Gas Law to solve for volume \( V \):
\[ V = \frac{nRT}{P} \]

### Step 3: Substitute the values into the equation
Substituting the known values:
- \( n = 0.1 \text{ moles} \)
- \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \)
- \( T = 273 \text{ K} \)
- \( P = 1 \text{ atm} \)

Now plug in the values:
\[ V = \frac{0.1 \text{ moles} \times 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \times 273 \text{ K}}{1 \text{ atm}} \]

### Step 4: Calculate the volume
Calculating the volume:
\[ V = 0.1 \times 0.0821 \times 273 = 2.24 \text{ L} \]

### Final Answer
The volume of \( CO_2 \) at STP is **2.24 liters**.

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