Question

Вычислите объем (н.у.) и массу оксида углерода (IV), которые выделятся при взаимодействии 150 г 60%-ного раствора уксусной кислоты с 10,6 г карбоната натрия, содержащего 5% примесей

Answer 1

2СH₃COOH + Na₂CO₃ =2 CH₃COONa +CO₂ + H₂O
m(CH₃COOH)=m(CH₃COOH р-ра) * W(CH₃COOH)
m(CH₃COOH)=150г*0.6=90г
n(CH₃COOH)=m(CH₃COOH)/M(CH₃COOH)
M(CH₃COOH)=60г/моль
n(CH₃COOH)= 90г / 60г/моль= 1.5моль
W(Na₂CO₃)=100%-5%=95%
m(Na₂CO₃)=10.6г*0.95=10.07г
n(Na₂CO₃)=m(Na₂CO₃)/M(Na₂CO₃)
M(Na₂CO₃)=106г/моль
n(Na₂CO₃)=10.07г /106г/моль=0.095моль
по уравнению химической реакции n(CH₃COOH):n(Na₂CO₃)=2:1
Na₂CO₃ в недостатке 
по уравнению химической реакции n(CO₂)=0.095моль
V(CO₂)=n(CO₂)*Vm
Vm = 22.4 дм³/моль
V(CO₂)=0.095моль* 22.4дм³/моль=2.13дм³
m(CO₂)=n(CO₂)*M(CO₂)
M(CO₂)=44г/моль
m(CO₂)=0.095моль* 44г/моль=4.18г

Answer 2

Calculation of Volume and Mass of Carbon Dioxide (CO2) Produced

To calculate the volume and mass of carbon dioxide (CO2) produced when 150 g of a 60% solution of acetic acid reacts with 10.6 g of sodium carbonate containing 5% impurities, we need to follow a few steps.

Step 1: Calculate the moles of acetic acid (CH3COOH) and sodium carbonate (Na2CO3) used in the reaction.

The molar mass of acetic acid (CH3COOH) is 60.05 g/mol. The molar mass of sodium carbonate (Na2CO3) is 105.99 g/mol.

The moles of acetic acid can be calculated using the formula:

moles of acetic acid = mass of acetic acid / molar mass of acetic acid

Substituting the given values: moles of acetic acid = 150 g / 60.05 g/mol = 2.498 mol

The moles of sodium carbonate can be calculated using the formula:

moles of sodium carbonate = mass of sodium carbonate / molar mass of sodium carbonate

Substituting the given values: moles of sodium carbonate = 10.6 g / 105.99 g/mol = 0.100 mol

Step 2: Determine the stoichiometry of the reaction.

The balanced chemical equation for the reaction between acetic acid and sodium carbonate is:

2 CH3COOH + Na2CO3 -> 2 CH3COONa + H2O + CO2

From the balanced equation, we can see that 2 moles of acetic acid react with 1 mole of sodium carbonate to produce 1 mole of carbon dioxide.

Step 3: Calculate the moles of carbon dioxide produced.

Since the stoichiometry of the reaction is 2:1 for acetic acid to sodium carbonate, the moles of carbon dioxide produced will be half the moles of acetic acid used.

moles of carbon dioxide = 0.5 * moles of acetic acid = 0.5 * 2.498 mol = 1.249 mol

Step 4: Calculate the volume of carbon dioxide at standard temperature and pressure (STP).

At STP, 1 mole of any ideal gas occupies 22.4 liters of volume.

volume of carbon dioxide = moles of carbon dioxide * 22.4 L/mol = 1.249 mol * 22.4 L/mol = 27.98 L

Therefore, the volume of carbon dioxide produced is approximately 27.98 liters.

Step 5: Calculate the mass of carbon dioxide produced.

The molar mass of carbon dioxide (CO2) is 44.01 g/mol.

mass of carbon dioxide = moles of carbon dioxide * molar mass of carbon dioxide mass of carbon dioxide = 1.249 mol * 44.01 g/mol = 54.99 g

Therefore, the mass of carbon dioxide produced is approximately 54.99 grams.

To summarize: - The volume of carbon dioxide produced is approximately 27.98 liters. - The mass of carbon dioxide produced is approximately 54.99 grams.

Please note that these calculations assume ideal conditions and do not take into account any side reactions or impurities in the reactants.