Question

Сколько литров оксида серы (4) выделится при взаимодействии 10 г сульфита натрия и 10 г соляной кислоты?

Answer 1

10g           10g                           X L
Na2SO3+HCL-->NaCL+H2O+SO2
126          36.5                          22.4
M(Na2SO3)=126g/mol
M(HCL)=36.5g/mol
Vm=22.4L/mol
n(Na2SO3)=m/M=10/126=0.079 mol
n(HCL)=m/M=10/36.5=0.274 mol
n(Na2SO3)<n(HCL)
10/126 = x/22.4
x=1.78 л
ответ 1.78 л

Answer 2

Calculation of the Amount of Sulfur Dioxide Produced

To calculate the amount of sulfur dioxide (SO2) produced when 10 grams of sodium sulfite (Na2SO3) reacts with 10 grams of hydrochloric acid (HCl), we need to determine the stoichiometry of the reaction and use it to convert the given masses to moles.

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

Na2SO3 + 2HCl → 2NaCl + H2O + SO2

From the balanced equation, we can see that 1 mole of sodium sulfite reacts with 2 moles of hydrochloric acid to produce 1 mole of sulfur dioxide.

Now, let's calculate the number of moles of sodium sulfite and hydrochloric acid:

- The molar mass of sodium sulfite (Na2SO3) is calculated as follows: - 2(22.99 g/mol) + 32.07 g/mol + 3(16.00 g/mol) = 126.04 g/mol - The number of moles of sodium sulfite is calculated as follows: - 10 g / 126.04 g/mol ≈ 0.0794 mol

- The molar mass of hydrochloric acid (HCl) is calculated as follows: - 1(1.01 g/mol) + 35.45 g/mol = 36.46 g/mol - The number of moles of hydrochloric acid is calculated as follows: - 10 g / 36.46 g/mol ≈ 0.274 mol

Since the reaction stoichiometry tells us that 1 mole of sodium sulfite reacts with 2 moles of hydrochloric acid to produce 1 mole of sulfur dioxide, we can conclude that the limiting reactant is sodium sulfite. This means that all of the sodium sulfite will react, and the amount of sulfur dioxide produced will be determined by the amount of sodium sulfite.

Therefore, the number of moles of sulfur dioxide produced is equal to the number of moles of sodium sulfite, which is approximately 0.0794 mol.

To convert the number of moles of sulfur dioxide to liters, we need to use the ideal gas law, which states that 1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP).

Therefore, the volume of sulfur dioxide produced is calculated as follows:

- Volume of sulfur dioxide = number of moles of sulfur dioxide × molar volume at STP - Volume of sulfur dioxide = 0.0794 mol × 22.4 L/mol ≈ 1.78 L

Therefore, approximately 1.78 liters of sulfur dioxide will be produced when 10 grams of sodium sulfite and 10 grams of hydrochloric acid react.

Please note that the above calculation assumes ideal conditions and does not take into account any potential side reactions or deviations from ideal gas behavior.