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alekssr [168]
1 year ago
14

Heteropoly tungstate supported on tantalum oxide: A highly active acid catalyst for the selective conversion of fructose to 5-hy

droxy methyl furfural
Chemistry
1 answer:
horsena [70]1 year ago
4 0

TPA catalysts based on tantalum oxide (Ta2O5) were produced and studied by FT-infrared, X-ray diffraction, Laser Raman, and temperature programmed ammonia desorption.

<h3>How did Lewis acidity induced heteropoly tungstate catalysts?</h3>

Heteropoly tungstate was synthesised and spread over tin oxide containing tantalum ions in its secondary structure. Different spectroscopic approaches were used to estimate the physical and chemical characteristics of the produced compounds. The inclusion of Ta ions in heteropoly tungstate resulted in the formation of new Lewis acidic sites. These samples were evaluated for their ability to catalyse the conversion of fructose to 5-ethoxymethylfurfural (EMF) and the selective etherification of 5-hydroxymethylfurfural (HMF) with ethanol. The catalyst with 30% active component on SnO2 had the greatest HMF etherification activity, yielding 90% 5-ethoxymethylfurfural in 45 minutes. The catalysts were also capable of converting fructose into EMF in a single pot with a 68% yield.

learn more about lewis acidity refer:

brainly.com/question/15220646

#SPJ4

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Answer:

* x_{CH_3OH}=0.0425

* \%m/m_{CH_3OH}=7.31\%

* m=2.46m

Explanation:

Hello,

In this case, for the mole fraction of methanol we use the formula:

x_{CH_3OH}=\frac{n_{CH_3OH}}{n_{CH_3OH}+n_{water}}

Thus, we compute the moles of both water (molar mass 18 g/mol) and methanol (molar mass 32 g/mol):

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Hence, mole fraction is:

x_{CH_3OH}=\frac{0.456mol}{0.456mol+10.3mol}\\\\x_{CH_3OH}=0.0425

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\%m/m_{CH_3OH}=\frac{m_{CH_3OH}}{m_{CH_3OH}+m_{water}}*100\%\\\\\%m/m_{CH_3OH}=\frac{14.6g}{14.6g+185g}*100\%\\\\\%m/m_{CH_3OH}=7.31\%

And the molality, considering the mass of water in kg (0.185 kg):

m=\frac{n_{CH_3OH}}{m_{water}} =\frac{0.456mol}{0.185kg}\\ \\m=2.46m

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