Fed-batch material balance: a reactor initially contains 2 L of medium with 1 g·L^-1 substrate. It is fed with substrate solution at 1 g·L^-1 at 1 L·h^-1 for 10 h. If after 10 h the reactor substrate concentration is 0.5 g·L^-1, what mass of substrate was consumed by the culture?

Introduction:
Fed-batch calculations rely on a running mass balance that accounts for initial inventory, cumulative feed addition, withdrawals (if any), and current reactor concentration. This question tests the ability to translate concentration-time information into total substrate consumption by the culture.

Given Data / Assumptions:

• Initial volume V0 = 2 L with C0 = 1 g·L^-1 ⇒ initial substrate mass = 2 g.
• Feed: C_f = 1 g·L^-1 at F = 1 L·h^-1 for 10 h ⇒ added mass = 10 g; volume added = 10 L.
• No harvest or bleed (volume strictly increases by feed).
• Final volume V_f = 2 + 10 = 12 L; final concentration C_f,out = 0.5 g·L^-1 ⇒ final mass = 6 g.

Concept / Approach:
Total mass consumed = (initial mass + mass in feed) − final mass in reactor. This directly applies the integral species balance over the fed-batch period without needing rates.

Step-by-Step Solution:
Step 1: Compute initial mass: m0 = 2 L * 1 g·L^-1 = 2 g.Step 2: Compute feed mass: m_in = 1 g·L^-1 * (1 L·h^-1 * 10 h) = 10 g.Step 3: Compute final mass: m_f = 12 L * 0.5 g·L^-1 = 6 g.Step 4: Consumption = (m0 + m_in − m_f) = (2 + 10 − 6) g = 6 g.

Verification / Alternative check:
A quick reasonableness check: despite substantial feed, the final concentration is lower than the feed, indicating net consumption; the computed 6 g used is consistent with the dilution and uptake.

Why Other Options Are Wrong:

• 2 g or 4 g: Underestimates; ignore the large feed volume.
• 8 g or 12 g: Overestimates; do not reconcile initial, feed, and final inventory correctly.

Common Pitfalls:
Forgetting the reactor volume increase, using concentrations instead of masses directly, or overlooking any withdrawals (none here).

Final Answer:
6g