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Tatiana [17]
3 years ago
7

what+is+the+theoretical+yield+of+ammonia(in+grams)+of+19.25+grams+of+nitrogen+gas+and+11.35+grams+of+hydrogen+gas+are+allowed+to

+react
Chemistry
1 answer:
kondor19780726 [428]3 years ago
7 0
We are given the amount of Nitrogen gas and hydrogen gas reacted to form ammonia:

N2 = 19.25 grams
H2 = 11.35 grams

Set-up a balanced chemical equation:

N2 + 3H2 ==> 2NH3

The theoretical amount of ammonia that will be produced from the given amounts is:

First, we need to determine the limiting reactant to serve as our basis for calculation.

number of moles / stoichiometric ratio 

N2 = 19.25  g/ 28 g/mol / 1 = 0.6875
H2 = 11.35 g/ 2 g/mol /3 = 1.89 

The limiting reactant is N2.

0.6875 moles N2 * (2 NH3/ 1 N2) * 17 g/mol NH3

The amount of NH3 produced is 23.375 grams of ammonia. <span />
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How many grams of Na2So4 would be formed if 0.75 moles of NaOH reacted?
Nata [24]

Answer:

\boxed{\text{53 g }}

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M_r:                                 142.04  

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1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.

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