Question

question 48
consider the following (incomplete) reaction. how many l of carbon dioxide can be produced at 301 k and 0.975 atm, if you begin with 10.89 g of glucose? (you do not need to balance it.)
c₆h₁₂o₆ → 2 c₂h₆o(l) + 2 co₂(g)
options: 1.421 l, 0.1121 l, 5.331 l, 3.081 l

Explanation:

Step1: Calculate moles of glucose

Molar mass of \( C_6H_{12}O_6 \) is \( 180.16 \, g/mol \). Moles of glucose \( n = \frac{mass}{molar\ mass} = \frac{10.89\ g}{180.16\ g/mol} \approx 0.06045\ mol \).

Step2: Determine moles of \( CO_2 \)

From the reaction \( C_6H_{12}O_6
ightarrow 2CH_3CH_2OH + 2CO_2 \), 1 mol glucose produces 2 mol \( CO_2 \). So moles of \( CO_2 \), \( n_{CO_2} = 2 \times 0.06045\ mol = 0.1209\ mol \).

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

\( R = 0.0821 \, L\cdot atm/(mol\cdot K) \), \( P = 0.975\ atm \), \( T = 301\ K \). Solve for \( V \): \( V = \frac{nRT}{P} = \frac{0.1209\ mol \times 0.0821\ L\cdot atm/(mol\cdot K) \times 301\ K}{0.975\ atm} \approx 3.08\ L \).

Answer:

3.08 L (corresponding to the option 3.08 L)