a
student ran the vanillin reduction as shown below. if 5.52 g of
vanillinwas used and...
a
student ran the vanillin reduction as shown below. if 5.52 g of
vanillinwas used and 3.267g of product was obtained. what is tge
tgeiretical 100% yeild. what is the actual yeild.
C= 12.01g/mol
H= 1.008 g/mol
O= 16.00g/mol
Solutions
Expert Solution
Ans.
Actual yield is 3.267 g,
Theoretical yield is 5.5931g ,p
yeild is 58.41%
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Vanillin Reduction with Sodium Borohydride to form Vanillyl
Alcohol
Started with 2.517g vanillin and 5.0mL NaBH4
Crude Product= 7.568g
Recrystallized Product= 2.297g
a) Calculate the %yield for both the crude and "fully"
synthesized product
b) Explain why this is likely not 100%. How could this be
improved
c) Draw mechanism
A student ran the following reaction in the laboratory at 445
K
: PCl5(g)-------> PCl3(g) + Cl2(g)
When she introduced 1.28 moles of PCl5(g) into a 1.00 liter
container, she found the equilibrium concentration of PCl5(g) to be
1.24 M.
Calculate the equilibrium constant, Kc, she obtained for this
reaction.
A student ran the following reaction in the laboratory at 684 K:
N2(g) + 3H2(g) 2NH3(g) When she introduced 3.26×10-2 moles of N2(g)
and 6.07×10-2 moles of H2(g) into a 1.00 liter container, she found
the equilibrium concentration of H2(g) to be 5.83×10-2 M. Calculate
the equilibrium constant, Kc, she obtained for this reaction.
Kc = __
-A student ran the following reaction in the laboratory at
1143 K:
2SO2(g) +
O2(g)
2SO3(g)
When she introduced 8.19×10-2 moles of
SO2(g) and
8.56×10-2 moles of
O2(g) into a 1.00 liter container, she
found the equilibrium concentration of
O2(g) to be
6.08×10-2 M.
Calculate the equilibrium constant, Kc, she obtained for
this reaction.
Kc =____________
-A student ran the following reaction in the laboratory at
363 K:
CH4(g) +
CCl4(g)
2CH2Cl2(g)
When she introduced 4.64×10-2 moles of
CH4(g) and...
A student ran the folllowing reaction in the laboratory at 661
K:
2NH3(g) N2(g) + 3H2(g)
When she introduced NH3(g) at a pressure of 0.597 atm into a
1.00 L evacuated container, she found the equilibrium partial
pressure of H2(g) to be 0.879 atm.
Calculate the equilibrium constant, Kp, she obtained for this
reaction.
A student ran the following reaction in the laboratory at 720 K:
H2(g) + I2(g) 2HI(g) When she introduced 0.189 moles of H2(g) and
0.218 moles of I2(g) into a 1.00 liter container, she found the
equilibrium concentration of I2(g) to be 6.05×10-2 M. Calculate the
equilibrium constant, Kc, she obtained for this reaction. Kc =
A student ran the following reaction in the laboratory at 291 K:
2CH2Cl2(g) CH4(g) + CCl4(g) When she introduced 6.63×10-2 moles of
CH2Cl2(g) into a 1.00 liter container, she found the equilibrium
concentration of CH2Cl2(g) to be 4.92×10-3 M. Calculate the
equilibrium constant, Kc, she obtained for this reaction.
Kc = _____
A student ran the following reaction in the laboratory at 610
K:
CO(g) + Cl2(g) COCl2(g)
When she introduced 0.183 moles of CO(g) and 0.211 moles of Cl2(g)
into a 1.00 liter container, she found the equilibrium
concentration of Cl2(g) to be 6.72×10-2 M.
Calculate the equilibrium constant, Kc, she obtained for this
reaction.
2.A student ran the following reaction in the laboratory at 546
K:
COCl2(g) CO(g) + Cl2(g)
When she introduced 0.854 moles of COCl2(g) into a 1.00...
1. A student ran the following reaction in the laboratory at
582 K:
CO(g) + Cl2(g)
COCl2(g)
When she introduced 0.380 moles of
CO(g) and 0.403 moles of
Cl2(g) into a 1.00 liter container, she
found the equilibrium concentration of
COCl2(g) to be 0.342
M.
Calculate the equilibrium constant, Kc, she obtained for
this reaction.
2. The equilibrium constant, Kc, for the following
reaction is 1.29×10-3 at
540 K.
COCl2(g) CO(g) +
Cl2(g)
When a sufficiently large sample of
COCl2(g)...
Shown below is an oxidation-reduction reaction in acidic
solution. Which of the following correctly balances the
reaction?
ClO−4+Rb→ClO−3+Rb+
Select the correct answer below:
A. ClO−4+Rb+2H+→ClO−3+Rb++H2O
B. 2ClO−4+Rb+4H+→2ClO−3+Rb++2H2O
C. ClO−4+2Rb+2H+→ClO−3+2Rb++H2O
D. ClO−4+3Rb+2H+→ClO−3+3Rb++H2O