Question

Какой объем водорода выделится при взаимодействии избытка магния с 250 г раствора соляной кислоты с массовой долей 23,36%?

Answer 1

Дано:
m(p-pa HCl) = 250 r
ω(HCl) = 23.36%=0,2336
V(H2)-?

m(HCl) = 250 r*0,2336 = 58,4 r
M(HCl) = 36,5 г/моль
n(HCl) = m/M = 58,4 r/36,5 г/моль = 1.6 моль

Mg + 2HCl = MgCl2 + H2↑
Из УХР следует, что n(H2) = 0.5n(HCl)
n(H2) = 0,5*1,6 моль = 0,8 моль
Vm = 22,4 л/моль
V(H2) = n*Vm = 0,8 моль*22,4 л/моль = 17,92 л
Ответ: 17,92 л

Answer 2

Calculation of Hydrogen Gas Volume

To calculate the volume of hydrogen gas released when an excess of magnesium reacts with 250 g of a solution of hydrochloric acid with a mass fraction of 23.36%, we need to use the stoichiometry of the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between magnesium and hydrochloric acid is:

Mg + 2HCl → MgCl2 + H2

From the equation, we can see that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.

To calculate the moles of hydrochloric acid, we need to convert the mass of the solution to moles. We can use the formula:

moles = mass / molar mass

The molar mass of hydrochloric acid (HCl) is approximately 36.5 g/mol.

Calculating the moles of hydrochloric acid:

moles of HCl = (mass of solution) / (molar mass of HCl)

moles of HCl = (250 g) / (36.5 g/mol) = 6.8493 mol

Since the reaction is 1:2 between magnesium and hydrochloric acid, the moles of hydrogen gas produced will be twice the moles of hydrochloric acid used.

moles of H2 = 2 * moles of HCl = 2 * 6.8493 mol = 13.6986 mol

Now, to calculate the volume of hydrogen gas, we can use the ideal gas law:

PV = nRT

Where: - P is the pressure of the gas (assumed to be constant) - V is the volume of the gas - n is the number of moles of the gas - R is the ideal gas constant (approximately 0.0821 L·atm/(mol·K)) - T is the temperature in Kelvin (assumed to be constant)

Since the pressure and temperature are assumed to be constant, we can rearrange the equation to solve for V:

V = (n * R * T) / P

Assuming standard temperature and pressure (STP) conditions of 1 atm and 273.15 K, we can calculate the volume of hydrogen gas:

V = (13.6986 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm

V ≈ 317.8 L

Therefore, approximately 317.8 liters of hydrogen gas will be released when an excess of magnesium reacts with 250 g of a solution of hydrochloric acid with a mass fraction of 23.36%.

Please note that this calculation assumes ideal conditions and may not account for all factors that could affect the reaction.