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Tatiana [17]
2 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]2 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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For the reaction
azamat

Answer:

Mass = 5.56 g

Explanation:

Given data:

Mass of Cl₂ = 4.45 g

Mass of NaCl produced = ?

Solution:

Chemical equation:

2Cl₂ + 4NaOH     →   3NaCl + NaClO₂ + 2H₂O

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 4.45 g/ 71 g/mol

Number of moles = 0.063 mol

Now we will compare the moles of Cl₂ with NaCl.

                  Cl₂         :         NaCl

                    2          :          3

                 0.063      :        3/2×0.063 =0.095 mol

Mass of NaCl:

Mass = number of moles × molar mass

Mass = 0.095 mol × 58.5 g/mol

Mass = 5.56 g

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2 years ago
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