Question

При спиртовом брожении глюкозы получили этанол,которой окислили до кислоты.На полученную кислоту подействовали избытком гидрокарбоната калия,при этом выделилось 8,96 л газа(н.у.)Определите массу глюкозы,подвергшуюся брожению, если выход спирта в этом процессе 80

Answer 1

$C_{6}H_{12}O_{6}=2C_{2}H_{5}OH+2CO_{2} C_{2}H_{5}OH=CH_{3} COOH CH_{3} COOH+KHCO3=CH_{3}COOK+H_{2}O+CO_{2} 1) n(CO_{2}) = 8.96/22.4=0.4 2) n(CO_{2}) = n(C_{2}H_{5}OH)=0.4 3) n(C_{6}H_{12}O_{6})= 0.4/0.8/2=0.25 4) m(C_{6}H_{12}O_{6})=180*0.25=45$

 

Answer 2

Calculation of Glucose Mass

To determine the mass of glucose that underwent fermentation, we need to use the given information about the volume of gas produced and the yield of alcohol.

Given: - Volume of gas produced: 8.96 L (at standard temperature and pressure, n.u.) - Alcohol yield: 80%

To calculate the mass of glucose, we can use the ideal gas law and stoichiometry.

1. Convert the volume of gas to moles using the ideal gas law: - The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. - At standard temperature and pressure (STP), the values are P = 1 atm and T = 273.15 K. - Rearranging the equation, we have n = PV / RT.

Using the given volume of gas (8.96 L) and converting it to moles: - n = (8.96 L) * (1 atm) / (0.0821 L·atm/(mol·K) * 273.15 K) - n ≈ 0.364 mol

2. Determine the number of moles of alcohol produced: - The balanced equation for the fermentation of glucose to alcohol is: C6H12O6 → 2 C2H5OH + 2 CO2 - From the equation, we can see that 1 mole of glucose produces 2 moles of alcohol. - Therefore, the number of moles of alcohol produced is twice the number of moles of glucose consumed.

Since the yield of alcohol is 80%, we can calculate the number of moles of glucose consumed: - Moles of glucose = (0.364 mol alcohol) / (2 * 0.80) - Moles of glucose ≈ 0.227 mol

3. Calculate the molar mass of glucose: - The molar mass of glucose (C6H12O6) is the sum of the atomic masses of its constituent elements. - The atomic masses are: C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol.

Molar mass of glucose = (6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol) Molar mass of glucose ≈ 180.18 g/mol

4. Calculate the mass of glucose: - Mass of glucose = Moles of glucose * Molar mass of glucose - Mass of glucose = 0.227 mol * 180.18 g/mol - Mass of glucose ≈ 40.95 g

Therefore, the mass of glucose that underwent fermentation is approximately 40.95 grams.