What is the correct equation for the reaction quotient of the following reaction? 2Fe3+(aq) + Zn(s) ⇌ 2Fe2+(aq) + Zn2+(aq)
1 answer:
Explanation:
A reaction quotient is defined as the ratio of concentration of products over reactants raised to the power of their stoichiometric coefficients.
A reaction quotient is denoted by the symbol Q.
For example,
The reaction quotient for this reaction is as follows.
Q =
[Zn] will be equal to 1 as it is present in solid state. Therefore, we don't need to write it in the reaction quotient expression.
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Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of require 3 moles of
Thus 2.8 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 1 mole of give = 2 moles of
Thus 2.8 moles of give =
of
Mass of
Theoretical yield of liquid iron = 313.6 g
Experimental yield = 288 g
Now we have to calculate the percent yield
Therefore, the percent yield is, 91.8%
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.
