What is the mass of glucose required to produce 44 g of `CO_(2)`, on complete combustion?

Options

A. 30 g
B. 45 g
C. 60 g
D. 22 g

Answer

To find the mass of glucose required to produce 44 g of \( CO_2 \) upon complete combustion, we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of glucose.
The combustion of glucose (\( C_6H_{12}O_6 \)) can be represented as:
\[ C_6H_{12}O_6 + O_2 \rightarrow CO_2 + H_2O \]
Balancing the equation gives:
\[ C_6H_{12}O_6 + 6 O_2 \rightarrow 6 CO_2 + 6 H_2O \]
This indicates that 1 mole of glucose produces 6 moles of \( CO_2 \).

### Step 2: Calculate the molar mass of glucose.
The molar mass of glucose (\( C_6H_{12}O_6 \)) is calculated as follows:
- Carbon (C): 12 g/mol × 6 = 72 g/mol
- Hydrogen (H): 1 g/mol × 12 = 12 g/mol
- Oxygen (O): 16 g/mol × 6 = 96 g/mol
Adding these together:
\[ 72 + 12 + 96 = 180 \text{ g/mol} \]
Thus, the molar mass of glucose is 180 g/mol.

### Step 3: Calculate the molar mass of carbon dioxide.
The molar mass of \( CO_2 \) is calculated as follows:
- Carbon (C): 12 g/mol × 1 = 12 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol
Adding these together:
\[ 12 + 32 = 44 \text{ g/mol} \]
Thus, the molar mass of \( CO_2 \) is 44 g/mol.

### Step 4: Determine the amount of glucose needed to produce 44 g of \( CO_2 \).
From the balanced equation, we know that 1 mole of glucose produces 6 moles of \( CO_2 \). Therefore, the relationship can be established as follows:
\[ \text{If } 1 \text{ mole of glucose (180 g) produces } 6 \text{ moles of } CO_2 (264 \text{ g}), \]
we can set up a proportion to find out how much glucose is needed to produce 44 g of \( CO_2 \):
\[ \frac{180 \text{ g glucose}}{264 \text{ g } CO_2} = \frac{x \text{ g glucose}}{44 \text{ g } CO_2} \]
Cross-multiplying gives:
\[ x \cdot 264 = 180 \cdot 44 \]
Calculating the right side:
\[ 180 \cdot 44 = 7920 \]
So,
\[ 264x = 7920 \]
Now, solving for \( x \):
\[ x = \frac{7920}{264} \approx 30 \text{ g} \]

### Conclusion
The mass of glucose required to produce 44 g of \( CO_2 \) is approximately **30 g**.