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1. How many liters of CO2 can be generated from the fermentation process involving 200.5 g C6H12O6? The process has been done in a room with a temperature of 28°C and a pressure of 1.02 atm.

C6H12O6 (aq) → 2C2H5OH(l) + 2CO2(g)


a. 52.0 L

b. 51.0 L

c. 53.0 L

d. 54.0 L

Sagot :

Romulosansolisidn

Answer:

Volume of CO₂ : V=53.9 L

Further explanation

In general, the gas equation can be written

PV=nRT

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08205 L.atm / mol K

T = temperature, Kelvin

Given

200.5 g C₆H₁₂O₆

Temperature = 28°C+273=301 K

Pressure = 1.02 atm

Required

Volume of CO₂

Solution

Reaction

C₆H₁₂O₆(aq) ⇒ 2 C₂H₅OH(l) + 2 CO₂ (g)

mol C₆H₁₂O₆ :

= mass : molar mass

= 200.5 g : 180.16 g/mol

= 1.113

From the equation, mol CO₂ :

= 2/1 x mol C₆H₁₂O₆

= 2/1 x 1.113

= 2.226

Ideal gas law :

V=nRT/P

V=2.226 x 0.08205 x 301 /1.02

V=53.9 L

c.53.0 L

Explanation:

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