Answer:
1. 24.7g of Br2
2. 19.9g of HCl
Explanation:Please see attachment for explanation
How many moles of fe2o3 will be produced from 18.0 g of fe assuming o2 is available in excess
1 answer:
Answer:
0.161moles
Explanation:
Given parameters:
Mass of Fe = 18g
Oxygen gas is in excess
Unknown:
Number of moles of Fe₂O₃ produced = ?
Solution:
To start with, let us write a chemically balanced equation before proceeding to understand the nuances of this problem.
4Fe + 3O₂ → 2Fe₂O₃
In the equation above above, 4 mole of iron combined with 3 moles of oxygen gas to 2 moles of Fe₂O₃.
In solving this problem, we can identify that Fe is the limiting reactant since we have been told oxygen gas is in excess. The suggests that the extent to which the product is formed and the reaction proceeds hinges on the amount of Fe we have.
It is best to work from the given, or known reactant to the unknown
The known in this scenario is the mass of Fe. Let us find the number of moles of this specie;
Number of moles of Fe =
Molar mass of Fe = 56g/mol
Number of moles = = 0.32mol
Using this known number of moles of Fe, we can relate it to that of the unknown amount of the product and obtain the number of moles.
4 moles of Fe produced 2 moles of Fe₂O₃
0.32 moles of Fe will produce = 0.161moles
You might be interested in
Answer:
As the molecular mass of given amine is 163 g/mol (a odd number) it means that this compound contains a odd number of Nitrogen atoms. We will first apply Rule of Thirteen to get the molecular formula.
Rule of Thirteen:
First divide the parent peak value by 13 as,
= 163 ÷ 13
= 12.53
Now, multiply 13 by 12,
= 13 × 12 (here, 12 specifies number of carbon atoms)
= 156
Now subtract 156 from 163,
= 163 - 156
= 7
Add 7 into 12,
= 7 + 12
= 19 (hydrogen atoms)
So, the rough formula we have is,
C₁₂H₁₉
Now, add one Nitrogen atom to above formula and subtract one Carbon and 2 Hydrogen atoms as these numbers are equal to atomic mass of Nitrogen atom as,
C₁₂H₁₉ -------N--------> C₁₁H₁₇N
Also, as shown in ¹³C-NMR there is one peak around 180 ppm and the peak at 1661 cm⁻¹ in IR spectrum is characteristic to carbonyl group hence, we will add one oxygen atom to the chemical formula accordingly. i.e.
C₁₁H₁₇N -------O--------> C₁₀H₁₃NO
Molecular Formula: C₁₀H₁₃NO
Also,
In NMR the the four peaks around 120 ppm are assigned to a mono substituted benzene ring.
The absence of IR peak above 3200 cm⁻³ also confirms that the amine is tertiary in nature and there is no hydrogen attached to the nitrogen atom.
It can be observed that the peaks in upfield are duplicating. This can be due to the presence of rotamers of said compound.
The most plausible structure for given data is shown below, and the resonance structure along with rotamers are also shown.
