Continuous reactor productivity from outlet composition and flow: the feed contains 40 g·L^-1 maltose; the reactor effluent contains 20 g·L^-1 lactate at a flow rate of 10 L·h^-1. What is the (mass) productivity of lactate leaving the reactor, based on the given data?

Introduction:
Productivity in a continuous process can refer to the mass of product discharged per unit time at the outlet flow. When reactor volume is not provided, one reports the mass flow rate of product based on outlet concentration and volumetric flow. This quick calculation is useful for comparing throughput across operating conditions.

Given Data / Assumptions:

• Outlet (effluent) lactate concentration = 20 g·L^-1.
• Volumetric flow rate F = 10 L·h^-1.
• No additional information about reactor volume; thus we compute mass per hour leaving in the effluent stream.

Concept / Approach:
Mass productivity as mass per unit time is simply the product of outlet concentration and volumetric flow rate: ṁ_product = C_product,out * F. Here it yields a numeric mass per hour. (Note: a volumetric productivity g·L^-1·h^-1 would require reactor volume to compute dilution rate D = F/V.)

Step-by-Step Solution:
Step 1: Identify C_lactate,out = 20 g·L^-1.Step 2: Identify F = 10 L·h^-1.Step 3: Compute mass flow: ṁ = 20 * 10 = 200 g·h^-1.Step 4: Report the corresponding choice consistent with the numeric value among options.

Verification / Alternative check:
Dimensional check confirms g·L^-1 * L·h^-1 = g·h^-1. The computed 200 g·h^-1 aligns with the throughput implied by the data.

Why Other Options Are Wrong:

• 50 g entries: Do not match the product C*F.
• 200 g maltose.l^-1: Refers to a different species and mismatched interpretation.
• 20 g lactate.h^-1: Off by a factor of 10 from the correct mass flow.

Common Pitfalls:
Confusing mass flow rate with volumetric productivity (g·L^-1·h^-1). Always check whether reactor volume is known before computing per-volume rates.

Final Answer:
200 g lactate.l^-1