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timurjin [86]
3 years ago
5

The reaction to prepare methanol from carbon monoxide and hydrogen ‍ is exothermic. If you wanted to use this reaction to produc

e methanol commercially, would high or low temperatures favor a maximum yield? Explain.
Chemistry
1 answer:
Alex_Xolod [135]3 years ago
8 0

Answer:

The reaction in low temperature will favor maximum yield.

Explanation:

Methanol is mainly produced by hydrogenation of carbon monoxide.It is the simplest alcohol which contain methyl group that is linked with hydroxyl group.Methanol is volatile,colorless,flammable liquid.

Ethanol is less toxic than methanol.Methanol is the precursor of acetic acid,methyl tertiary butyl ether.

During formation of commercial methanol,carbon dioxide and water acts as poisons to alkali metal methoxide catalyst in low temperature in methanol synthesis reaction.

More over,high temperature evaporates methanol.

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Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 5.54-g sample of vinegar was neutralized by 30.10 mL of 0.1
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Answer:

3.26 % of vinegar is acetic acid

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Step 1: Data given

Mass of the sample = 5.54 grams

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Molarity of NaOH = 0.100M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles NaOH

Moles NaOH = volume * molarity

Moles NaOH = 0.03010 L * 0.100M

Moles NaOH = 0.00301 moles NaOH

Step 4: Calculate moles CH3COOH

For 1 mol NaOH we need 1 mol CH3COOH

For 0.00301 moles NaOH we nee 0.00301 moles CH3COOH

Step 5: Calculate mass CH3COOH

Mass CH3COOH = moles CH3COOH * molar mass CH3COOH

Mass CH3COOH = 0.00301 moles * 60.05 g/mol

Mass CH3COOH = 0.1808 grams

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