Question

При прокаливании 18,36 г смеси KNO3,AgNO3,Cu(NO3)2 объем выделенных газов составил 4.032 л (н.у.).Твердый осадок обработали водой, после чего его масса уменьшелась на 3.7г.Найдите массовые доли солей в исходной смеси. помогите пожалуйста

Answer 1

2KNO3=KNO2+O2
2AgNO3=2Ag+2NO2+O2
Cu(NO3)2=CuO+2NO2+O2
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Потом нитрит калия растворился. Его-то и получилось 3,4 г
М (KNO2)=39+14+2*16=85
3,4/85=0,04 моля
НитрАта калия было вдвое больше 0,08 моль или 0,08*(39+14+3*16)=8,08 г
Остальные нитраты весили 10,28 г
Нитрат калия выделил при разложении 0,04*22,4=0,896 л газа
Пусть нитрата серебра было х молей, а нитрата меди - у
(108+14+3*16)х+(64+14+3*16)у=10,28
170х+126у=10,28
По газу
1,5x+3y=(4,032-0,896)/22,4
{85x+63y=5,14
{3x+6y=0,28
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x=(0,28-6y)/3
85(0,28-6y)+189y=15.42
23,8-510y+189y=15,42
321y=8,38, y=0,026 моля или 3,28 г нитрата меди
нитрата серебра было 10,28-3,28=7,00 г
Массовые доли
KNO3 8,08/18,36=....%
Cu(NO3)2 3,28/18,36=....%
AgNO3 7/18,36=...%
3

Answer 2

Calculation of Mass Fractions of Salts in the Mixture

To find the mass fractions of the salts in the initial mixture, we need to use the given information about the volume of gas evolved and the change in mass of the solid precipitate.

Let's denote the mass of the mixture as M (in grams) and the mass of each salt as m1, m2, and m3 (in grams) for KNO3, AgNO3, and Cu(NO3)2, respectively.

According to the given information, the volume of gas evolved during the reaction is 4.032 L (at standard temperature and pressure). This volume can be converted to the number of moles of gas using the ideal gas law.

Using the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature, we can calculate the number of moles of gas evolved.

Since the volume is given at standard temperature and pressure (STP), we can use the values: - Pressure (P) = 1 atm - Temperature (T) = 273.15 K - Ideal gas constant (R) = 0.0821 L·atm/(mol·K)

Let's assume the number of moles of gas evolved is n.

Now, let's consider the change in mass of the solid precipitate. The mass of the precipitate before treatment with water is M - 3.7 g, and after treatment, it becomes M.

From the given information, we can set up the following equations:

Equation 1: n = V / (RT) Equation 2: M - 3.7 g = m1 + m2 + m3

To find the mass fractions of the salts, we need to express the mass of each salt in terms of the total mass M.

Let's assume the mass fraction of KNO3 is x1, AgNO3 is x2, and Cu(NO3)2 is x3.

We can write the following equations:

Equation 3: m1 = x1 * M Equation 4: m2 = x2 * M Equation 5: m3 = x3 * M

Substituting Equations 3, 4, and 5 into Equation 2, we get:

M - 3.7 g = x1 * M + x2 * M + x3 * M

Simplifying the equation, we have:

M - 3.7 g = (x1 + x2 + x3) * M

Now, we can solve the equations to find the mass fractions of the salts.

Let's proceed with the calculations.