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il63 [147K]
3 years ago
8

Medicinal “ether" is produced when ethyl alcohol is treated with an acid. How many grams of

Chemistry
1 answer:
DochEvi [55]3 years ago
6 0

Answer:

34.3 g

Explanation:

Step 1: Write the balanced equation

2 CH₃CH₂OH ⇒ CH₃CH₂OCH₂CH₃ + H₂O

Step 2: Calculate the moles corresponding to 50.0 g of CH₃CH₂OH

The molar mass of CH₃CH₂OH is 46.07 g/mol.

50.0 g × 1 mol/46.07 g = 1.09 mol

Step 3: Calculate the theoretical moles of CH₃CH₂OCH₂CH₃ produced

The molar ratio of CH₃CH₂OH to CH₃CH₂OCH₂CH₃ is 2:1. The moles of CH₃CH₂OCH₂CH₃ theoretically produced are 1/2 × 1.09 mol = 0.545 mol.

Step 4: Calculate the real moles of CH₃CH₂OCH₂CH₃ produced

The percent yield of the reaction is 85%.

0.545 mol × 85% = 0.463 mol

Step 5: Calculate the mass corresponding to 0.463 moles of CH₃CH₂OCH₂CH₃

The molar mass of CH₃CH₂OCH₂CH₃ is 74.12 g/mol.

0.463 mol × 74.12 g/mol = 34.3 g

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Consider the reaction 2 NO + O2 → 2 NO2
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Answer:

(a) Rate at which NO_2 is formed is 0.050 M/s

(b) Rate at which O_2 is consumed is 0.0250 M/s.

Explanation:

The given reaction is:-

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The expression for rate can be written as:-

-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}

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-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}

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\frac{d[NO_2]}{dt}=0.050\ M/s

(a) Rate at which NO_2 is formed is 0.050 M/s

-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}

-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}

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(b) Rate at which O_2 is consumed is 0.0250 M/s.

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3 years ago
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Answer:

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