Answer:
Moles of H₂S needed = 6.2 mol
Moles of SO₂ produced = 6.2 mol
Explanation:
Given data:
Number of moles of O₂ = 9.3 mol
Moles of H₂S needed = ?
Moles of SO₂ produced = ?
Solution:
Chemical equation:
2H₂S + 3O₂ → 2SO₂ + 2H₂O
Now we will compare the moles of oxygen with H₂S.
O₂ : H₂S
3 : 2
9.3 : 2/3×9.3 = 6.2 mol
Now we will compare the moles of SO₂ with both reactant.
O₂ : SO₂
3 : 2
9.3 : 2/3×9.3 = 6.2 mol
H₂S : SO₂
2 : 2
6.2 : 6.2 mol
So 6.2 moles of SO₂ are produced.
Answer:
See explanation.
Explanation:
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In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:
Thus, we proceed as follows:
Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:
Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:
Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:
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Explanation:
Combustion of a compound is the reaction with oxygen , hence , the process of combustion is an oxidation reaction.
The carbohydrates contain more amount of oxygen as compared to the fats ,
Hence ,
carbohydrates , have a lot of oxygen contents , are are already partially oxidized , but fats have lower oxygen content .
Therefore ,
The partially oxidized carbohydrates are very difficult to oxidized in comparison to fats .
Rutherford's Gold Foil Experiment proved the existence of a small massive center to atoms, which would later be known as the nucleus of an atom. Ernest Rutherford, Hans Geiger and Ernest Marsden carried out their Gold Foil Experiment to observe the effect of alpha particles on matter.
Answer:
Explanation:
From the net ionic equation
Ba2+(aq) + SO42-(aq) ==> BaSO4(s) we see that 1 mole Ba2+ reacts with 1 mole SO42- to -> 1 mol BaSO4
Find moles of Ba2+ used: 0.250 moles/L x 0.0323 L = 0.008075 moles Ba2+
Find moles SO42- present: 0.008075 moles Ba2+ x 1 mol SO42-/1 mol Ba2+ = 0.008075 mol SO42-
Find mass of Na2SO4 present: 0.008075 mol SO42- x 1 mol Na2SO4/1 mol SO42- x 142.04 Na2SO4/mole = 1.14698 g = 1.15 g Na2SO4 (to 3 significant figures)
