Is sponge cake a heterogeneous mixture
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The mass of sulfurous acid produced when 245g of sulfur dioxide is reacted with water is 313.8 g
<h3>Stoichiometry </h3>From the question, we are to determine the mass of sulfurous acid produced
First, we will write the balaced chemical equation for the reaction
H₂O + SO₂ → H₂SO₃
This means 1 mole of H₂O reacts with 1 mole of sulfur dioxide (SO₂) to produce 1 mole of sulfurous acid (H₂SO₃)
Now, we will determine the number of moles of sulfur dioxide present
Mass of sulfur dioxide present = 245 g
Molar mass of sulfur dioxide = 64.066 g/mol
Using the formula,
Then,
Number of moles of SO₂ present =
Number of moles of SO₂ present = 3.824 moles
This is the number of moles of SO₂ present
From balanced chemical equation
1 mole of H₂O reacts with 1 mole of sulfur dioxide (SO₂) to produce 1 mole of sulfurous acid (H₂SO₃)
Then,
Water (H₂O) will react with the 3.824 moles of sulfur dioxide (SO₂) to produce 3.824 moles of sulfurous acid (H₂SO₃)
∴ The number of moles of sulfurous acid produced is 3.824 moles
Now, for the mass of sulfurous acid produced
Molar mass of sulfurous acid = 82.07 g/mole
Using the formula,
Mass = Number of moles × Molar mass
Mass of sulfurous acid produced = 3.824 × 82.07
Mass of sulfurous acid produced = 313.8 g
Hence, the mass of sulfurous acid produced is 313.8 g
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The limiting reactant is lead(II) acetate the theoretical yield is 1.1909 g and the percent yield is 83.71 %.
<h3>What is a molarity ?</h3>The amount of the sample in a specific solution volume is known as its molarity (M). The molarity of a solute per liter of the a solution is known as molar ratio. The molar concentration in a solution is another name for molarity.
<h3>Briefing:</h3>Molarity
= moles of solute/volume of the solution
or Moles
= Molarity * volume of solution
For potassium sulfate :
Molarity = 0.116 M
Volume = 56.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 56.0×10⁻³ L
Thus, moles of potassium sulfate:
Moles = 0.116 M * 56.0 * 10-3 moles
Moles of potassium sulfate = 0.006496 moles
For lead(II) acetate :
Molarity = 0.102 M
Volume = 38.5 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 38.5×10⁻³ L
Thus, moles of lead(II) acetate :
Moles = 0.102 * 38.5 *10-3 moles
Moles of lead(II) acetate = 0.003927 moles
According to the given reaction:
1 mole of potassium sulfate react with 1 mole of lead(II) acetate
0.006496 moles potassium sulfate react with 0.006496 mole of lead(II) acetate
Moles of lead(II) acetate = 0.003927 moles
Lead(II) acetate is limiting reagent. ( 0.003927 < 0.006496)
0.003927 mole of lead(II) acetate gives 0.003927 mole of lead(II) sulfate
Molar mass of lead(II) sulfate = 303.26 g/mol
Mass of lead(II) sulfate = Moles × Molar mass = 0.003927 × 303.26 g = 1.1909 g
Theoretical yield = 1.1909 g
Given experimental yield = 0.997 g
% yield = (Experimental yield / Theoretical yield) × 100 = (0.997/1.1909 g) × 100 = 83.71 %
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