What mass of ammonia can be produced if 13.4 grams of nitrogen gas reacted ?

Chemistry

1 answer:

aivan3 [116]3 years ago

3 0

Answer:

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

Explanation:

Step 1: Data given

Mass of nitrogen gas (N2) = 13.4 grams

Molar mass of N2 = 28 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate moles of N2

Moles N2 = Mass N2 / molar mass N2

Moles N2 = 13.4 grams / 28.00 g/mol

Moles N2 = 0.479 moles

Step 4: Calculate moles of NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles

Step 5: Calculate mass of NH3

Mass of NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.958 moles * 17.03 g/mol

Mass NH3 = 16.3 grams

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

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lukranit [14]

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3 years ago

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NISA [10]

Answer:

Explanation:

The movement of the electrons is illustrated in the picture attached to this answer. It is a four-step reaction mechanism.

First STEP: The first step involves the transfer of an electron from sodium to form a radical anion.

Second STEP: This radical anion then removes a proton/hydrogen from ammonia in a bid to neutralize itself (hence the hydrogen becomes bonded to the anion).

Third STEP: The sodium (from NaNH₂ formed) transfers an electron again to produce a vinyl carbanion.

Fourth STEP: The carbanion then removes a proton/hydrogen from ammonia (like in the second step) to form a neutral trans-alkene.

NOTE: The circled numbers denote each step while the mechanism on the left represents the use of any alkyl group (R and R') while the mechanism on the right assumes both alkyl groups are methyl. Hence, 2-butyne started the reaction and the final product was trans-2-butene.