GATE 2019 BT Q.51: solution and answer

GATE 2019 · BTQ.51 · 2 marks · NAT

Yeast biomass ( $C_{6} H_{10} O_{3} N$ ) grown on glucose is described by the stoichiometric equation given below:

C_{6} H_{12} O_{6} + 0.48 NH_{3} + 3 O_{2} \to 0.48 C_{6} H_{10} O_{3} N + 3.12 CO_{2} + 4.32 H_{2} O

The amount of glucose needed for the production of $50 g L^{- 1}$ of yeast biomass in a batch reactor with a working volume of 1,00,000 L is ________ kg (rounded off to the nearest integer).

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Answer and solution

ANSWER

About 13021 kg of glucose is needed.

Official answer: 12500 to 14500Source: Official answer key, GATE 2019 BT (IIT Madras)

Total biomass to be produced: $m_{X} = 50 g L^{- 1} \times 1, 00, 000 L = 5 \times 1 0^{6} g = 5000 kg$ .
Molar masses (C = 12, H = 1, O = 16, N = 14): biomass $C_{6} H_{10} O_{3} N$ : $72 + 10 + 48 + 14 = 144 g mol^{- 1}$ ; glucose $C_{6} H_{12} O_{6}$ : $72 + 12 + 96 = 180 g mol^{- 1}$ .
From the stoichiometric equation, 1 mol of glucose gives 0.48 mol of biomass. The biomass yield on substrate, $Y_{X / S}$ (grams of biomass formed per gram of glucose consumed), is therefore $Y_{X / S} = \frac{0.48 \times 144}{1 \times 180} = \frac{69.12}{180} = 0.384 g biomass (g glucose)^{- 1}$
Glucose required: $m_{S} = \frac{m _{X}}{Y _{X / S}} = \frac{5000 kg}{0.384} = 13020.8 kg \approx 13021 kg$
Check by moles: $n_{X} = \frac{5 \times 1 0 ^{6} g}{144 g mol ^{- 1}} = 34722.2 mol$ ; $n_{S} = \frac{34722.2}{0.48} = 72338.0 mol$ ; $m_{S} = 72338.0 \times 180 g = 1.30208 \times 1 0^{7} g \approx 13021 kg$ .

KEY IDEA

The coefficients in a balanced growth equation are in moles. Multiply each one by its molar mass to get a mass yield: $Y_{X / S} = (0.48 \times 144) /180 = 0.384 g g^{- 1}$ . The glucose needed is then the total biomass divided by $Y_{X / S}$ . The equation balances for C, H, O and N, so its coefficients can be used as printed. Only the biomass and glucose coefficients matter here; NH3, O2, CO2 and H2O play no part in the answer.

WATCH OUT FOR

Treating 0.48 as a mass ratio, which gives $5000/0.48 \approx 10417$ kg (outside the key), or forgetting to convert the molar ratio to a mass ratio with the two molar masses (144 and 180). Another slip is misreading the volume: 1,00,000 L in Indian digit grouping is $1 0^{5}$ L. Reading it as $1 0^{6}$ L makes every figure ten times too large.

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GATE 2019 BT Q.51: solution and answer

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