Explanation:
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The balanced equation for the reaction is as follows;
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂
But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant.
H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced
Answer:
Reducing agent - S
Explanation:
The equation for the reaction is given as;
K2Cr2O3 + H2O + S --> SO2 + KOH + Cr2O3
The balanced equation is given as:
2K2Cr2O7 + 2H2O + 3S → 3SO2 + 4KOH + 2Cr2O3
The reducing agent can be obtained by comparing the oxidation number of the elements. Increase in oxidation number signifies a reducing agent
K:
Reactant - 1
Product - 1
Cr
Reactant - +2
Product - +3
O
Reactant - -2
Product - -2
H
Reactant - +1
Product - +1
S
Reactant - O
Product - +4
S has an increase in oxidation number hence it is the reducing agent.
Although Cr has an increase in oxidation number, it occurs as a compound.
Answer:
The answer is
<h2>38.85 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
density = 0.3456 g/cm³
volume = 112.4 cm³
The mass of the object is
mass = 0.3456 × 112.4 = 38.84544
We have the final answer as
<h3>38.85 g</h3>
Hope this helps you
According to
Graham's Law of Diffusion,"the rates of diffusion of two gases are inversely proportional to the square root of their Molar masses or Densities at the same pressure and temperature".
r₁ / r₂ =
Where,
r₁ = Rate of Fluorine
r₂ = Rate of Bromine
M₂ = Molar mass of Bromine = 159.8 g/mol
M₁ = Molar mass of Fluorine = 37.98 g/mol
Putting values,
r₁ / r₂ =
r₁ / r₂ =
r₁ / r₂ =
2.04Result: Fluorine effuses
2 times faster than Bromine gas.