The ratio of the volumes of 11 g of `CO_(2)` and 28 g of CO at STP is __________ .

Answer

To solve the problem of finding the ratio of the volumes of 11 g of CO₂ and 28 g of CO at STP, we can follow these steps:

### Step 1: Calculate the number of moles of CO₂
To find the number of moles of CO₂, we use the formula:

\[ n = \frac{W}{M} \]

Where:
- \( W \) = mass of the gas (11 g for CO₂)
- \( M \) = molar mass of CO₂

The molar mass of CO₂ is calculated as follows:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol (and there are 2 oxygen atoms in CO₂)

So, the molar mass of CO₂ is:

\[ M_{CO₂} = 12 + (16 \times 2) = 12 + 32 = 44 \text{ g/mol} \]

Now, substituting the values into the formula:

\[ n_{CO₂} = \frac{11 \text{ g}}{44 \text{ g/mol}} = \frac{1}{4} \text{ mol} \]

### Step 2: Calculate the number of moles of CO
Next, we calculate the number of moles of CO using the same formula:

\[ n_{CO} = \frac{W}{M} \]

Where:
- \( W \) = mass of the gas (28 g for CO)
- \( M \) = molar mass of CO

The molar mass of CO is calculated as follows:
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol

So, the molar mass of CO is:

\[ M_{CO} = 12 + 16 = 28 \text{ g/mol} \]

Now, substituting the values into the formula:

\[ n_{CO} = \frac{28 \text{ g}}{28 \text{ g/mol}} = 1 \text{ mol} \]

### Step 3: Calculate the volumes of CO₂ and CO at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters.

- Volume of CO₂:

\[ V_{CO₂} = n_{CO₂} \times 22.4 \text{ L/mol} = \left(\frac{1}{4} \text{ mol}\right) \times 22.4 \text{ L/mol} = 5.6 \text{ L} \]

- Volume of CO:

\[ V_{CO} = n_{CO} \times 22.4 \text{ L/mol} = 1 \text{ mol} \times 22.4 \text{ L/mol} = 22.4 \text{ L} \]

### Step 4: Calculate the ratio of the volumes
Now, we can find the ratio of the volumes of CO₂ to CO:

\[ \text{Ratio} = \frac{V_{CO₂}}{V_{CO}} = \frac{5.6 \text{ L}}{22.4 \text{ L}} = \frac{1}{4} \]

### Final Answer
The ratio of the volumes of 11 g of CO₂ and 28 g of CO at STP is:

\[ \frac{1}{4} \]