Question

consider the following (incomplete) reaction. how many l of carbon dioxide can be produced at 325 k and 1.12 atm, if you begin with 10.89 g of glucose (c₆h₁₂o₆)? (you do not need to balance it.)
c₆h₁₂o₆ → 2 c₅h₅oh (l) + 2 co₂ (g)
options:
○ 1.44 l
○ 2.88 l
○ 5.32 l
○ 259 l

Explanation:

Step1: Calculate moles of glucose

Molar mass of \( C_6H_{12}O_6 \) is \( 6\times12.01 + 12\times1.008 + 6\times16.00 = 180.16 \, \text{g/mol} \).
Moles of glucose \( n = \frac{10.89 \, \text{g}}{180.16 \, \text{g/mol}} \approx 0.06045 \, \text{mol} \).

Step2: Relate moles of glucose to \( CO_2 \)

From the reaction, \( 1 \, \text{mol} \, C_6H_{12}O_6 \) produces \( 2 \, \text{mol} \, CO_2 \).
Moles of \( CO_2 \), \( n_{CO_2} = 2 \times 0.06045 \, \text{mol} = 0.1209 \, \text{mol} \).

Step3: Apply ideal gas law (\( PV = nRT \))

\( R = 0.0821 \, \text{L·atm/(mol·K)} \), \( T = 325 \, \text{K} \), \( P = 1.12 \, \text{atm} \).
Solve for \( V \): \( V = \frac{nRT}{P} = \frac{0.1209 \, \text{mol} \times 0.0821 \, \text{L·atm/(mol·K)} \times 325 \, \text{K}}{1.12 \, \text{atm}} \).
Calculate: \( V \approx \frac{0.1209 \times 0.0821 \times 325}{1.12} \approx 2.88 \, \text{L} \).

Answer:

2.88 L