Question

96г смеси карбоната и гидрокарбоната натрия прокалили до постоянной массы, равной 69г. Какова масса компонентов в исходной смеси?

Answer 1

При нагревании разложился только гидрокарбонат натрия.
2NaHCO3=Na2CO3+H2O+CO2
M(NaHCO3)=84г/моль
М(H2O и CO2)=18+44=62г/моль
Смесь стала легче за счёт удаления воды и углекислого газа.
Ушло 96г-69г=27г
Составим пропорцию по уравнению реакции и узнаем, сколько гидрокарбоната надо разложить, чтобы выделилось 27г воды и углекислого газа.
из ХгNaHCO3 выделится 27г
а из 168г/моль - 62 г/моль
Х=73г
Значит исходная смесь содержала 73гNaHCO3, и 96-73=23г Na2CO3

Answer 2

Calculation of the Mass of Components in the Initial Mixture

To calculate the mass of the components in the initial mixture, we need to use the information provided. According to the question, a mixture of sodium carbonate and sodium hydrocarbonate was heated until a constant mass of 69g was obtained. The initial mass of the mixture is not given.

To solve this problem, we can set up an equation based on the conservation of mass. Let's assume the initial mass of the mixture is x grams. The equation can be written as:

x grams (initial mass) = 69 grams (final mass)

Now, we need to determine the mass of the components in the initial mixture. Let's assume the mass of sodium carbonate is y grams and the mass of sodium hydrocarbonate is z grams.

According to the equation, we can write:

y grams (mass of sodium carbonate) + z grams (mass of sodium hydrocarbonate) = x grams (initial mass)

Since we don't have the values of y and z, we cannot directly calculate their masses. However, we can use the molar masses of sodium carbonate (Na2CO3) and sodium hydrocarbonate (NaHCO3) to set up a system of equations.

The molar mass of sodium carbonate (Na2CO3) is approximately 105.99 g/mol The molar mass of sodium hydrocarbonate (NaHCO3) is approximately 84.01 g/mol.

Let's assume the number of moles of sodium carbonate is n1 and the number of moles of sodium hydrocarbonate is n2.

We can write the following equations:

n1 * 105.99 g/mol + n2 * 84.01 g/mol = y grams (mass of sodium carbonate)

n1 * 105.99 g/mol + n2 * 84.01 g/mol = z grams (mass of sodium hydrocarbonate)

Since we have two equations and two unknowns, we can solve this system of equations to find the values of y and z.

Unfortunately, the specific values needed to solve this problem are not provided in the search results. Therefore, I am unable to provide the exact mass of the components in the initial mixture.