Question

M(C12H22O11)=120 г V(CO2)-? СРОЧНООО

Answer 1

С12Н22О11+12О2=12СО2+11Н2О

n(C12H22O11)=120г/342 г/моль=0,35 моль

По УХР СО2 в 12 раз больше сахарозы, тогда n(CO2)=0,35*12=4,2 моль

V(CO2)=4,2 моль*22,4моль/л=94,08 л

Answer 2

To calculate the volume of carbon dioxide (CO2) produced from the combustion of the sugar molecule (C12H22O11), you first need to balance the chemical equation for the combustion reaction.

The balanced chemical equation for the combustion of glucose (C6H12O6) can be used as a reference since it is also a sugar molecule, and it follows a similar reaction pattern. The balanced equation for the combustion of glucose is:

C6H12O6 + 6O2 → 6CO2 + 6H2O

Now, let's write the balanced equation for the combustion of sucrose (C12H22O11):

C12H22O11 + 12O2 → 12CO2 + 11H2O

From the balanced equation, we can see that 1 mole of C12H22O11 produces 12 moles of CO2. The molar mass of C12H22O11 (sucrose) is approximately 342.3 g/mol.

Now, we can use this information to calculate the volume of CO2 produced:

Step 1: Calculate the number of moles of C12H22O11 (sucrose) using its mass: Molar mass of C12H22O11 = 342.3 g/mol Mass of C12H22O11 = 120 g

Number of moles = Mass / Molar mass Number of moles = 120 g / 342.3 g/mol ≈ 0.3503 mol

Step 2: Calculate the number of moles of CO2 produced (using the balanced equation): According to the balanced equation, 1 mole of C12H22O11 produces 12 moles of CO2.

Number of moles of CO2 = Number of moles of C12H22O11 × 12 Number of moles of CO2 = 0.3503 mol × 12 ≈ 4.2036 mol

Step 3: Calculate the volume of CO2 at standard conditions (STP): At standard conditions (STP: 0°C and 1 atm), 1 mole of any ideal gas occupies 22.71 L.

Volume of CO2 at STP = Number of moles of CO2 × 22.71 L/mol Volume of CO2 at STP = 4.2036 mol × 22.71 L/mol ≈ 95.41 L

So, approximately 95.41 liters of CO2 would be produced from 120 grams of sucrose (C12H22O11) under standard conditions.