Answer:
14.3 g SO₃
Explanation:
2S + 3O₂ → 2SO₃
First, find the limiting reactant. To do that, calculate the mass of oxygen needed to react with all the sulfur.
5.71 g S × (1 mol S / 32 g S) = 0.178 mol S
0.178 mol S × (3 mol O₂ / 2 mol S) = 0.268 mol O₂
0.268 mol O₂ × (32 g O₂ / mol O₂) = 8.57 g O₂
There are 10.0 g of O₂, so there's enough oxygen. The limiting reactant is therefore sulfur.
Use the mass of sulfur to calculate the mass of sulfur trioxide.
5.71 g S × (1 mol S / 32 g S) = 0.178 mol S
0.178 mol S × (2 mol SO₃ / 2 mol S) = 0.178 mol SO₃
0.178 mol SO₃ × (80 g SO₃ / mol SO₃) = 14.3 g SO₃
Answer:
<u> </u><u>85.952 ml</u> needed to titrate the excess complexing reagent .
Explanation:
Lets calculate
After addition of 80 ml of EDTA the solution becomes = 20 + 70 = 90 ml
As the number of moles of =
=
=0.01172
Molarity =
=
=0.000586 moles
Excess of EDTA = concentration of EDTA - concentration of CoSO4
= 0.009005 - 0.000586
= 0.008419 M
As M1V1 ( Excess of EDTA ) = M2V2
V2 =85.952 ml
Therefore , <u>85.952 ml </u> needed to titrate the excess complexing reagent .
Your independent variable would be salt because it doesn't change the only thing that changes is the water.
There are<u> </u><u>11.5 moles</u> of potassium contained in 449 g fundamental units of potassium. therefore,
option B is correct.
A mole is the mass of a substance made up of the same number of fundamental components. Atoms in a 12-gram example are identical to 12C. Depending on the substance, the fundamental units may be molecules, atoms, or formula units.
Atomic weight 39.1 is assigned to potassium (chemical symbol K).
K content in a 449g sample is equal to
=given weight / atomic weight
=449/39.1
= 11.5mol.
A 449g sample has 11.5 moles of potassium.
Learn more about fundamental units here-
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