One electron has a mass of 1 amu 2 amu 0 amu 3 amu
1 answer:
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Answer:
0,051g of O₂
Explanation:
The reaction of precipitation of Fe is:
4Fe(OH)⁺(aq) + 4OH⁻(aq) + O₂(g) + 2H₂O(l) → 4Fe(OH)₃(s)
<em>-Where the Fe(OH)⁺ is Fe(II) and Fe(OH)₃ is Fe(III)-</em>
This reaction is showing you need 1 mol of O₂(g) per 4 moles of Fe(II) for a complete reaction.
85mL= 0,085L of 0,075M Fe(II) are:
0,085L*0,075M = 6,34x10⁻³ moles of Fe(II)
For a complete reaction of 6,34x10⁻³ moles of Fe(II) you need:
6,34x10⁻³ moles of Fe(II)× =
1.59x10⁻³ moles of O₂. In grams:
1.59x10⁻³ moles of O₂× = <em>0,051g of O₂</em>
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I hope it helps!
Answer:- B: is the right answer.
Solution:- The balanced equation is:
We have been given with 8.75 grams of oxygen and asked to calculate the grams of hydrogen needed to react with given grams of oxygen according to the balanced equation.
From balanced equation, 1 mole of oxygen reacts with 2 moles of hydrogen.
So, let's convert grams of oxygen to moles and multiply it by the mole ratio to calculate the moles of hydrogen that are easily converted to grams on multiplying by it's molar mass.
The complete set up looks as:
=
Hence, the right option is B: .