Answer:
6 moles of Oxygen required
Answer:
6.0 x 10^-1 g Fe
Explanation:
I might be in the wrong but hear me out:
Convert 0.85 g to mols of Fe2O3.
0.85 g x 1 mol/159.69=0.0053 mols Fe2O3.
Get moles of Fe (iron metal).
0.0053 mols Fe2O3 x 2 mol Fe/ 1 mol Fe2O3
= 1.0 x 10^-2 mols Fe
Convert moles to grams of Fe.
1.0 x 10^-2 mols of Fe x 55.85 grams Fe/ 1 mol Fe
= 6.0 x 10^-1
There should be only one sig fig. Hoped this helped.
Answer:
Mass FeCl2 = 0.0333g
Mass CrCl2 = 0.9961g
Explanation:
To solve this problem. The first equation we can write is:
Mass FeCl2 + Mass CrCl2 = 1.0294g <em>(1)</em>
Now, the Chlorides of FeCl2 and CrCl2 react producing 2.3609g of AgCl
Using molar mass of these species (126.75g/mol, 122.9g/mol, 143.32g/mol, respectively), you can write the equation:
2Mass FeCl2 / 126.75 + 2Mass CrCl2 / 122.9 = 2.3609/143.32
<em>That is: Moles Chloride before = Moles Chloride in AgCl after reaction</em>
7.8895x10⁻³Mass FeCl2 + 0.0162734MassCrCl2 = 0.016473 <em>(2)</em>
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Replacing (1) in (2):
7.8895x10⁻³ (1.0294g - MassCrCl2) + 0.0162734MassCrCl2 = 0.016473
8.12145x10⁻³ -7.8895x10⁻³MassCrCl2 + 0.0162734Mass CrCl2 = 0.016473
8.3839x10⁻³ MassCrCl2 = 8.35155x10⁻³
Mass CrCl2 = 0.9961g
And:
Mass FeCl2 = 1.0294g - 0.9961g
Mass FeCl2 = 0.0333g
2 electrons on the first ring, 8 on the second, and 3 on the third ring
Hey there!
N₂H₄ + O₂ → NO₂ + H₂O
First, balance N.
Two on the left, one on the right. Add a coefficient of 2 in front of NO₂.
N₂H₄ + O₂ → 2NO₂ + H₂O
Next, balance H.
Four on the left, two on the right. Add a coefficient of 2 in front of H₂O.
N₂H₄ + O₂ → 2NO₂ + 2H₂O
Lastly, balance O.
Two on the left, six on the right. Add a coefficient of 3 in front of O₂.
N₂H₄ + 3O₂ → 2NO₂ + 2H₂O
This is our final balanced equation.
Hope this helps!