`896 mL`. of a mixture of ` CO `and `CO_2` weigh `1.28 g` at NTP. Calculate the volume of `CO_2` in the mixture at NTP.

Options

A. `448 ml `
B. `672 ml`
C. `224 ml`
D. `500 ml`

Answer

To solve the problem of finding the volume of \( CO_2 \) in a mixture of \( CO \) and \( CO_2 \) that weighs 1.28 g and occupies 896 mL at NTP, we can follow these steps:

### Step 1: Understand the Given Data
- Total volume of the gas mixture = 896 mL
- Total weight of the gas mixture = 1.28 g
- Molar mass of \( CO \) = 28 g/mol
- Molar mass of \( CO_2 \) = 44 g/mol

### Step 2: Set Up the Equations
Let the weight of \( CO \) be \( x \) grams. Then, the weight of \( CO_2 \) will be \( 1.28 - x \) grams.

### Step 3: Calculate Moles of Each Gas
- Moles of \( CO \) = \( \frac{x}{28} \)
- Moles of \( CO_2 \) = \( \frac{1.28 - x}{44} \)

### Step 4: Total Moles in the Mixture
The total volume of the gas mixture at NTP can be converted to moles using the molar volume of a gas at NTP, which is 22,400 mL/mol.

\[ \text{Total moles} = \frac{896 \text{ mL}}{22400 \text{ mL/mol}} = 0.04 \text{ moles} \]

### Step 5: Set Up the Equation for Total Moles
Now, we can set up the equation for the total moles:

\[ \frac{x}{28} + \frac{1.28 - x}{44} = 0.04 \]

### Step 6: Solve for \( x \)
To solve this equation, we first find a common denominator, which is 28 * 44 = 1232.

Multiplying through by 1232 to eliminate the denominators:

\[ 44x + 28(1.28 - x) = 0.04 \times 1232 \]

Calculating \( 0.04 \times 1232 = 49.28 \):

\[ 44x + 35.84 - 28x = 49.28 \]

Combining like terms:

\[ 16x + 35.84 = 49.28 \]

Subtracting 35.84 from both sides:

\[ 16x = 49.28 - 35.84 \]
\[ 16x = 13.44 \]
\[ x = \frac{13.44}{16} = 0.84 \text{ g} \]

### Step 7: Calculate the Weight of \( CO_2 \)
Now, we can find the weight of \( CO_2 \):

\[ \text{Weight of } CO_2 = 1.28 - x = 1.28 - 0.84 = 0.44 \text{ g} \]

### Step 8: Calculate Moles of \( CO_2 \)
Now, we can calculate the moles of \( CO_2 \):

\[ \text{Moles of } CO_2 = \frac{0.44}{44} = 0.01 \text{ moles} \]

### Step 9: Calculate the Volume of \( CO_2 \)
Finally, we can find the volume of \( CO_2 \) at NTP:

\[ \text{Volume of } CO_2 = 0.01 \text{ moles} \times 22400 \text{ mL/mol} = 224 \text{ mL} \]

### Final Answer
The volume of \( CO_2 \) in the mixture at NTP is **224 mL**.

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