Which of the following combination of volumes (mL) of `C_2H_4`(g) and `O_2`(g) respectively is required to obtain 400 mL of `CO_2`(g)? All volumes are measured at STP.

Options

A. 100 , 300
B. 200 , 600
C. 200 , 200
D. 300 , 300

Answer

To solve the problem of determining the volumes of ethylene (`C2H4`) and oxygen (`O2`) required to produce 400 mL of carbon dioxide (`CO2`), we can follow these steps:

### Step 1: Write the balanced chemical equation
The combustion of ethylene can be represented by the following balanced equation:

\[ C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(g) \]

### Step 2: Determine the stoichiometric ratios
From the balanced equation, we can see the stoichiometric ratios of the reactants and products:
- 1 volume of `C2H4` reacts with 3 volumes of `O2` to produce 2 volumes of `CO2`.

### Step 3: Relate the volume of `CO2` to the volumes of `C2H4` and `O2`
We know from the problem that we want to obtain 400 mL of `CO2`. Since the stoichiometry tells us that 2 volumes of `CO2` are produced from 1 volume of `C2H4`, we can calculate the volume of `C2H4` needed:

\[ \text{Volume of } C_2H_4 = \frac{1}{2} \times \text{Volume of } CO_2 = \frac{1}{2} \times 400 \, \text{mL} = 200 \, \text{mL} \]

### Step 4: Calculate the volume of `O2` required
Using the stoichiometric ratio, we can now determine the volume of `O2` required. Since 1 volume of `C2H4` requires 3 volumes of `O2`:

\[ \text{Volume of } O_2 = 3 \times \text{Volume of } C_2H_4 = 3 \times 200 \, \text{mL} = 600 \, \text{mL} \]

### Step 5: Summarize the results
Thus, to produce 400 mL of `CO2`, we require:
- Volume of `C2H4` = 200 mL
- Volume of `O2` = 600 mL

### Conclusion
The correct combination of volumes of `C2H4` and `O2` required to obtain 400 mL of `CO2` is:
- **200 mL of `C2H4` and 600 mL of `O2`.**