Question

Ребята срооооооочноооо!! Какую массу (NH4)2SO4 можно получить при взаимодействии H2SO4 с 1 кг аммиака? Какой объём 60%-го раствора H2SO4 (p=1,5 г/см³) будет при этом затрачено ???

Answer 1

Дано:
m(NH3)=1 кг = 1000 г
ω(H2SO4) = 60% или 0,6
р=1,5 г/см³
Найти:
m((NH4)2SO4)=?
V(p-pa H2SO4)=?

Решение.
M((NH4)2)SO4)= 132 г/моль
M(NH3) = 17 г/моль
n(NH3) = m(NH3)/M(NH3)
n(NH3) = 1000 г/17 г/моль = 58,82 моль
M(H2SO4)= 98 г/моль

2NH3 + H2SO4 = (NH4)2SO4
Из УХР следует, что
n(NH4)2SO4)=n(H2SO4)=1/2n(NH3) = 1/2*58.82=29.41 моль
m(NH4)2SO4) = n(NH4)2SO4)*M(NH4)2SO4)
m(NH4)2SO4)= 29,41 моль*132 г/моль = 3882,12 г≈3,9 кг
m(H2SO4) = n(H2SO4)*M(H2SO4)
m(H2SO4) = 29.41 моль*98 г/моль = 2882,18 г
m(p-pa H2SO4) = m(H2SO4)/ω(H2SO4)
m(p-pa H2SO4) = 2882,18 г/0,6 = 4803,63 г
V(p-pa H2SO4) = m(p-pa H2SO4)/p(H2SO4)
V(p-pa H2SO4) = 4803.63 г/1,5 г/см³ = 3202,42 см³≈3,2 дм³
Ответ: 3,9 кг (NH4)2SO4); 3,2 дм³ p-pa H2SO4

Answer 2

Calculating the Mass of (NH4)2SO4 Produced

To calculate the mass of (NH4)2SO4 produced from the reaction of H2SO4 with 1 kg of ammonia (NH3), we can use the balanced chemical equation for the reaction and the stoichiometry of the reaction.

The balanced chemical equation for the reaction is: 2 NH3 + H2SO4 → (NH4)2SO4

From the balanced equation, we can see that 2 moles of ammonia react with 1 mole of sulfuric acid to produce 1 mole of ammonium sulfate.

Given that the molar mass of ammonia (NH3) is approximately 17 g/mol and the molar mass of sulfuric acid (H2SO4) is approximately 98 g/mol, we can calculate the amount of (NH4)2SO4 produced.

1. First, we need to convert the mass of ammonia (1 kg) to moles: - 1 kg = 1000 g - Moles of NH3 = Mass / Molar mass = 1000 g / 17 g/mol

2. Next, we use the mole ratio from the balanced equation to find the moles of (NH4)2SO4 produced: - Moles of (NH4)2SO4 = (Moles of NH3) / 2

3. Finally, we convert the moles of (NH4)2SO4 to grams: - Mass of (NH4)2SO4 = Moles * Molar mass

Let's calculate the mass of (NH4)2SO4 produced using these steps.

Calculation: 1. Moles of NH3 = 1000 g / 17 g/mol = 58.82 moles 2. Moles of (NH4)2SO4 = 58.82 moles / 2 = 29.41 moles 3. Mass of (NH4)2SO4 = 29.41 moles * (molar mass of (NH4)2SO4)

Calculating the Volume of 60% H2SO4 Solution Required

To calculate the volume of 60% H2SO4 solution required for the reaction, we can use the concept of molarity and the given density of the solution.

The molarity (M) of a solution is defined as the number of moles of solute per liter of solution. We can use the equation: Molarity (M) = (moles of solute) / (volume of solution in liters)

Given that the density (p) of the 60% H2SO4 solution is 1.5 g/cm³, we can use the density and the percent by mass to calculate the molarity of the solution.

Calculation: 1. Calculate the mass of H2SO4 in the solution: - Mass of H2SO4 = 60% * Volume of solution * Density of solution

2. Convert the mass of H2SO4 to moles using the molar mass of H2SO4: - Moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4

3. Use the molarity equation to find the volume of solution required: - Volume of solution = (moles of H2SO4) / Molarity

Let's calculate the volume of 60% H2SO4 solution required using these steps.

Note: The molar mass of H2SO4 is approximately 98 g/mol.

I will now calculate the mass of (NH4)2SO4 produced and the volume of 60% H2SO4 solution required based on the provided information.