if 15 mol C is mixed with 10 mol O2 which reactant is the limiting reactant? Which reactant would be the excess reactant?
1 answer:
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Answer:
The boiling point is 308.27 K (35.27°C)
Explanation:
The chemical reaction for the boiling of titanium tetrachloride is shown below:
Ti ⇒ Ti
ΔH° (Ti
) = -804.2 kJ/mol
ΔH° (Ti
) = -763.2 kJ/mol
Therefore,
ΔH° = ΔH°
(Ti
) - ΔH°
(Ti
) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol
Similarly,
s°(Ti) = 221.9 J/(mol*K)
s°(Ti) = 354.9 J/(mol*K)
Therefore,
s° = s° (Ti) - s°(Ti
) = 354.9 - 221.9 = 133 J/(mol*K)
Thus, T = ΔH° /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C
Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.
Answer:
A. 10.0 grams of ethyl butyrate would be synthesized.
B. 57.5% was the percent yield.
C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.
Explanation:
A
Moles of butanoic acid =
According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :
of ethyl butyrate
Mass of 0.08636 moles of ethyl butyrate =
0.08636 mol × 116 g/mol = 10.0 g
Theoretical yield = 10.0 g
Experimental yield = ?
Percentage yield of the reaction = 100%
Experimental yield = 10.0 g
10.0 grams of ethyl butyrate would be synthesized.
B
Theoretical yield of ethyl butyrate = 10.0 g
Experimental yield ethyl butyrate = 5.75 g
Percentage yield of the reaction = ?
57.5% was the percent yield.
C
Moles of butanoic acid =
According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :
of ethyl butyrate
Mass of 0.08636 moles of ethyl butyrate =
0.08636 mol × 116 g/mol = 10.0 g
Theoretical yield = 10.0 g
Experimental yield = ?
Percentage yield of the reaction = 78.0%
Experimental yield = 7.80 g
7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.