Question

При взаимодействии оксида серы (6 валентной )с водой образовалась серная кислота массой 720г,выход продукта составляет 82%.Определите массу оксида серы израсходованой во время реакции

Answer 1

SO₃ + H₂O = H₂SO₄

m'(H₂SO₄)=720 г
w=0,82
M(H₂SO₄)=98 г/моль
M(SO₃)=80 г/моль

практическая масса серной кислоты
m'(H₂SO₄)=wm(H₂SO₄)

теоретическая масса серной кислоты
m(H₂SO₄)=m'(H₂SO₄)/w

по уравнению реакции
m(SO₃)/M(SO₃)=m(H₂SO₄)/M(H₂SO₄)

масса оксида серы (VI)
m(SO₃)=M(SO₃)m'(H₂SO₄)/{wM(H₂SO₄)}

m(SO₃)=80*720/{0,82*98}=717 г

Answer 2

Calculation of the Mass of Sulfur Dioxide Reacted

To determine the mass of sulfur dioxide (SO2) reacted during the reaction, we can use the given information about the mass of sulfuric acid formed and the yield of the reaction.

Let's assume that the mass of sulfur dioxide reacted is "x" grams.

According to the balanced chemical equation for the reaction between sulfur dioxide and water to form sulfuric acid:

SO2 + H2O -> H2SO4

The molar mass of sulfuric acid (H2SO4) is 98.09 g/mol, and the molar mass of sulfur dioxide (SO2) is 64.06 g/mol.

Given that the mass of sulfuric acid formed is 720 g and the yield of the reaction is 82%, we can set up the following equation:

(82/100) * (98.09 g/mol) * x = 720 g

Simplifying the equation:

(0.82) * (98.09 g/mol) * x = 720 g

Solving for x:

x = (720 g) / [(0.82) * (98.09 g/mol)]

Calculating the value of x:

x ≈ 9.74 mol

To convert the moles of sulfur dioxide to grams, we can use the molar mass of sulfur dioxide:

Mass of sulfur dioxide = (9.74 mol) * (64.06 g/mol)

The mass of sulfur dioxide reacted during the reaction is approximately 623.45 grams.