Question

Кальций карбонат взаимодействует с соляной кислотой. Вычислите, какую массу соли нужно взять для получения карбон (IV) оксида объёмом 33,6 л (н. у.). Практический выход продукта 85%.

Answer 1

Ответ:

Дано:

V(CO2)=28.56л

Vпракт(CO2)=85%=0.85

Найти:

m(Na2Co3)-?

Решение:

Na2CO3+2HCl=2NaCl+CO2+H2O

1) Находим теоретический объем (СO2), по формуле (V/Vпракт)

28.56л/0.85= 33.6л

2) Находим ню СО2 по формуле (V/V молярный)

33.6/22.4=1.5 моль

3) Смотря на коэфф. ур-ия, мы видим, что у нас взяты 1 моль Na2CO3 и 1 моль СО2. Из этого следует, что

ню(Na2CO3)=ню(СО2)=1.5моль

4) Прежде чем найти массу карбоната натрия, нам надо найти его молярную массу (смотря на табл. Менделеева)

М (Na2CO3) = 2*23+12+3*16=106г/моль

5) Теперь мы можем найти то, что нас требуют по усл. задачи.

m= ню* М (молярная масса)

m(Na2CO3) = 1.5моль* 106г/моль = 159г

Ответ: m(Na2CO3) = 159 грамм

Answer 2

Calculation of the Mass of Calcium Carbonate Needed

To calculate the mass of calcium carbonate needed to obtain 33.6 liters of carbon (IV) oxide at standard conditions (n.u.) with a practical product yield of 85%, we need to follow a few steps.

1. Determine the molar ratio between calcium carbonate and carbon (IV) oxide. 2. Calculate the number of moles of carbon (IV) oxide. 3. Use the molar ratio to find the number of moles of calcium carbonate. 4. Convert the moles of calcium carbonate to grams.

Let's go through these steps one by one.

Step 1: Determine the Molar Ratio

The balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl) to produce carbon (IV) oxide (CO2) is as follows:

CaCO3 + 2HCl -> CaCl2 + CO2 + H2O

From the equation, we can see that 1 mole of calcium carbonate reacts to produce 1 mole of carbon (IV) oxide.

Step 2: Calculate the Number of Moles of Carbon (IV) Oxide

To calculate the number of moles of carbon (IV) oxide, we need to use the ideal gas law equation:

PV = nRT

Where: P = pressure (atm) V = volume (L) n = number of moles R = ideal gas constant (0.0821 L.atm/mol.K) T = temperature (K)

At standard conditions (n.u.), the pressure is 1 atm and the temperature is 273 K. Therefore, we have:

PV = nRT (1 atm) * (33.6 L) = n * (0.0821 L.atm/mol.K) * (273 K)

Solving for n, we find:

n = (1 atm * 33.6 L) / (0.0821 L.atm/mol.K * 273 K)

Step 3: Use the Molar Ratio

Since the molar ratio between calcium carbonate and carbon (IV) oxide is 1:1, the number of moles of calcium carbonate needed is the same as the number of moles of carbon (IV) oxide.

Step 4: Convert Moles to Grams

To convert the moles of calcium carbonate to grams, we need to use the molar mass of calcium carbonate. The molar mass of calcium carbonate is calculated by adding up the atomic masses of its constituent elements: calcium (Ca), carbon (C), and oxygen (O).

The atomic masses are as follows: - Calcium (Ca): 40.08 g/mol - Carbon (C): 12.01 g/mol - Oxygen (O): 16.00 g/mol

The molar mass of calcium carbonate (CaCO3) is: 40.08 g/mol (Ca) + 12.01 g/mol (C) + (16.00 g/mol (O) * 3) = 100.09 g/mol

Finally, we can calculate the mass of calcium carbonate needed by multiplying the number of moles of calcium carbonate by its molar mass:

Mass = moles * molar mass

Now, let's perform the calculations.

Calculation:

Step 1: Determine the Molar Ratio - The molar ratio between calcium carbonate and carbon (IV) oxide is 1:1.

Step 2: Calculate the Number of Moles of Carbon (IV) Oxide - Using the ideal gas law equation: - P = 1 atm - V = 33.6 L - R = 0.0821 L.atm/mol.K - T = 273 K - n = (1 atm * 33.6 L) / (0.0821 L.atm/mol.K * 273 K) - Calculate the value of n.

Step 3: Use the Molar Ratio - The number of moles of calcium carbonate needed is the same as the number of moles of carbon (IV) oxide.

Step 4: Convert Moles to Grams - The molar mass of calcium carbonate (CaCO3) is 100.09 g/mol. - Calculate the mass of calcium carbonate needed by multiplying the number of moles by the molar mass.

Please wait a moment while I perform the calculations.