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3 years ago

In the following reaction, how many grams of ammonia (NH3) will produce

2 answers:

3 0

145.899 grams

Take the 300 grams of N2 you want and divide by its molar mass of 17.0337 to tell you that you want 10.70664 moles of N2. Since for every 5 moles of N2 produced, you need 4 moles of ammonia, divide by 5 and then multiply by 4 to get the number of moles of ammonia you need. This will be 10.70664 / 5 * 4 = 8.56531 moles. Finally multiply by the molar mass of ammonia and you'll know how many grams of ammonia you need. This is 8.56531 * 17.0337 = 145.8989 grams of ammonia which rounds off to 145.899 grams.

lara [203]3 years ago

3 0

Answer:

The correct option is the third option which is 145.899g

Explanation:

From the equation of reaction: 4NH3 + 6NO → 5N2 + 6H2O

First calculate the masses involved in the given equation of reaction.

Mass of ammonia NH3 = 4 x 17.0337= 68.1348

Mass of nitrogen N2 = 5 x 28. 02 = 140.1

Use the following analogy

68.1348g of ammonia produced 140.1g of N2

Yg of ammonia will be produced by 300g of N2 (Using Y as unknown mass)

Calculating for Y by cross-multiplying 

Therefore 300g of N2 (Y) will be produced by 68.1348 x 300/ 140.1 = 145.899g

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4 years ago

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Rank the species CH2NH, CH3I, HF, C2H6 from most to least miscible in ammonia (NH3) 1. CH3I > C2H6 > HF > CH2NH 2. CH2N

777dan777 [17]

Answer:

$HF>CH_{2}NH>CH_{3}I>C_{2}H_{6}$

Explanation:

$NH_{3}$ is an weak base. HF is an weak acid. Hence an acid-base reaction will ocuur between $NH_{3}$ and HF resulting a homogeneous mixture.

$NH_{3}$ is a polar protic molecule. Hence it can form hydrogen bonding with another polar protic molecule $CH_{2}NH$ . Also dipole-dipole attraction force along with london dispersion force exists between $CH_{2}NH$ and $NH_{3}$ .

$CH_{3}I$ is a polar molecule. Hence only dipole-dipole attraction force along with london dispersion force operates between $CH_{3}I$ and $NH_{3}$ .

$C_{2}H_{6}$ is a non-polar molecule. Hence only london dispersion force exists between $C_{2}H_{6}$ and $NH_{3}$ .

So, order of miscibility: $HF>CH_{2}NH>CH_{3}I>C_{2}H_{6}$

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3 years ago

A drop of vinegar ( a weak acid ) is placed on a sample of each of the materials below. Which will show the most active reaction

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Answer: Option (B) is the correct answer.

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The reaction is as follows.

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3 years ago

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