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xlameweirdox
20.11.2019 •
Chemistry
Consider the reaction
. how much nh₃ can be produced from the reaction of 74.2 g of n₂ and 14.0 moles of h₂? a. 1. 1.59 × 10²⁴ molecules b. 2. 1.69 × 10²⁵ molecules c. 3. 3.19 × 10²⁴ molecules d. 4. 1.26 × 10²⁵ molecules e. 5. 5.62 × 10²⁴ molecules
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Ответ:
E
How much NH₃ can be produced from the reaction below:
N₂ + 3H₂ - 2NH₃
The stoichiometric ratio of the reactants = 1:3
Given
74.2g of N₂, and Molar mass = 14g/mole
Mole of N₂ = 74.2/14=5.3mols of N₂,
and 14mols of H₂
From this given values and comparing with the stoichiometric ratio, H₂ will be the limiting reagent while N₂ is the excess reactant.
i.e, for every 14mols of H₂, we need 4.67mols of N₂ to react with it to produce 9.33mols of NH₃ as shown (vice versa)
From this we have 9.33mols of NH₃ produced
Avogadro constant, we have n = no of particles = 6.022x10²³ molecules contained in every mole of an element.
For a 9.33mols of NH3, we have 9.33x6.022x10²³molecules in NH3
5.62x10²⁴molecules of NH₃
Ответ: