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 determine the mass of glucose (C₆H₁₂O₆) required to produce 44 g of CO₂ through complete combustion, we can follow these steps:

### Step 1: Write the balanced chemical equation for the combustion of glucose.
The balanced equation for the complete combustion of glucose is:
\[ C_6H_{12}O_6 + 6 O_2 \rightarrow 6 CO_2 + 6 H_2O \]

### Step 2: Determine the molar mass of glucose (C₆H₁₂O₆).
To find the molar mass of glucose:
- Carbon (C): 6 atoms × 12 g/mol = 72 g/mol
- Hydrogen (H): 12 atoms × 1 g/mol = 12 g/mol
- Oxygen (O): 6 atoms × 16 g/mol = 96 g/mol

Adding these together:
\[ \text{Molar mass of glucose} = 72 + 12 + 96 = 180 \text{ g/mol} \]

### Step 3: Determine the amount of CO₂ produced from glucose.
From the balanced equation, 1 mole of glucose produces 6 moles of CO₂. Therefore, we can set up a ratio:
- 1 mole of C₆H₁₂O₆ produces 6 moles of CO₂.
- Molar mass of CO₂ = 44 g/mol.

### Step 4: Calculate the mass of glucose needed to produce 44 g of CO₂.
First, find out how many moles of CO₂ are in 44 g:
\[ \text{Moles of } CO_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{44 \text{ g}}{44 \text{ g/mol}} = 1 \text{ mole} \]

Since 1 mole of CO₂ is produced from \(\frac{1}{6}\) moles of glucose:
\[ \text{Moles of glucose required} = \frac{1 \text{ mole of } CO_2}{6} = \frac{1}{6} \text{ moles of glucose} \]

### Step 5: Convert moles of glucose to grams.
To find the mass of glucose required:
\[ \text{Mass of glucose} = \text{moles} \times \text{molar mass} = \frac{1}{6} \text{ moles} \times 180 \text{ g/mol} = 30 \text{ g} \]

### Final Answer:
The mass of glucose required to produce 44 g of CO₂ is **30 g**.

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