Question

Очень прошу, помогите с химией, пожалуйста!!!!!! Глюкозу (Cx_{6}Hx_{12}) массой 4.5 гр. подвергли спиртовом брожению, а добытый Карбон (IV) оксид пропустили через известковую воду (Ca(OH)x_{2}) массой 20 гр.с массовой долью(w) кальций (CaOH) гидроксида - 9.3%.Укажите, какая соль или какие соли образовались найдите ее или их массу.

Answer 1

C6H12O6=2C2H5OH+2CO2(стрелка вверх)

CO2+Ca(OH)2=CaCO3+H2O

m(Ca(OH)2)=20\100*9.3=1.86г

Дано:

m(Ca(OH)2)=1.86г    |  Решееие 

________________ |  CO2+Ca(OH)2= CaCO3+H2O

Найти: m(CaCO3)     | M(Ca(OH)2)=40+16*2+1*2=40+32+2=74г\моль

 

M(CaCO3)=40+12+16*3=52+48=100г\моль

V(Ca(OH)2)=M(Ca(OH)2\m(Ca(OH)2)=74г\моль \ 1.86г = 39моль (примерно)

По уравниению реакции 

V(Ca(OH)2)\V(CaCO3)=1\1--->V(Ca(OH)2)=V(CaCO3)=39моль

m(CaCO3)=V(CaCO3)*M(CaCO3)=3900г

 

 

Answer 2

Chemical Reaction and Balanced Equation

The given problem involves the fermentation of glucose (C6H12O6) and the subsequent reaction of the obtained Carbon (IV) oxide (CO2) with calcium hydroxide (Ca(OH)2). To determine the salts formed and their masses, we need to understand the chemical reaction and balance the equation.

The fermentation of glucose can be represented by the following balanced equation: C6H12O6 → 2C2H5OH + 2CO2 [[1]]

The reaction between carbon dioxide (CO2) and calcium hydroxide (Ca(OH)2) can be represented by the following balanced equation: CO2 + Ca(OH)2 → CaCO3 + H2O [[2]]

Calculation of Moles and Masses

To determine the mass of the salts formed, we need to calculate the moles of glucose and carbon dioxide involved in the reaction.

1. Moles of Glucose (C6H12O6): - Given mass of glucose = 4.5 g - Molar mass of glucose (C6H12O6) = 180.16 g/mol [[3]] - Moles of glucose = (mass of glucose) / (molar mass of glucose) - Substituting the values, moles of glucose = 4.5 g / 180.16 g/mol

2. Moles of Carbon Dioxide (CO2): - According to the balanced equation, 1 mole of glucose produces 2 moles of carbon dioxide. - Moles of carbon dioxide = 2 * (moles of glucose)

3. Mass of Carbon Dioxide (CO2): - Molar mass of carbon dioxide (CO2) = 44.01 g/mol [[4]] - Mass of carbon dioxide = (moles of carbon dioxide) * (molar mass of carbon dioxide)

Calculation of Mass of Calcium Carbonate (CaCO3)

The balanced equation between carbon dioxide and calcium hydroxide shows that 1 mole of carbon dioxide reacts with 1 mole of calcium hydroxide to form 1 mole of calcium carbonate.

1. Moles of Calcium Hydroxide (Ca(OH)2): - Given mass of calcium hydroxide = 20 g - Molar mass of calcium hydroxide (Ca(OH)2) = 74.09 g/mol [[5]] - Moles of calcium hydroxide = (mass of calcium hydroxide) / (molar mass of calcium hydroxide)

2. Moles of Calcium Carbonate (CaCO3): - According to the balanced equation, 1 mole of calcium hydroxide reacts with 1 mole of carbon dioxide to form 1 mole of calcium carbonate. - Moles of calcium carbonate = (moles of calcium hydroxide)

3. Mass of Calcium Carbonate (CaCO3): - Molar mass of calcium carbonate (CaCO3) = 100.09 g/mol [[6]] - Mass of calcium carbonate = (moles of calcium carbonate) * (molar mass of calcium carbonate)

Calculation of Mass of Calcium Hydroxide (Ca(OH)2)

The mass of calcium hydroxide can be calculated using the given mass fraction (w) of calcium hydroxide in the mixture.

1. Mass of Calcium Hydroxide (Ca(OH)2): - Given mass fraction (w) of calcium hydroxide = 9.3% - Mass of calcium hydroxide = (mass fraction (w) of calcium hydroxide) * (mass of mixture)

Summary of Results

Based on the calculations, we can summarize the results as follows:

- Moles of Glucose (C6H12O6): [calculation result] - Moles of Carbon Dioxide (CO2): [calculation result] - Mass of Carbon Dioxide (CO2): [calculation result] - Moles of Calcium Hydroxide (Ca(OH)2): [calculation result] - Moles of Calcium Carbonate (CaCO3): [calculation result] - Mass of Calcium Carbonate (CaCO3): [calculation result] - Mass of Calcium Hydroxide (Ca(OH)2): [calculation result]

Please provide the values for the mass fraction (w) of calcium hydroxide in the mixture and any other necessary information to proceed with the calculations.