weirdoal567
26.07.2019 •
Chemistry
A3.82-g sample of magnesium nitride is reacted with 7.73 g of water. mg3n2 (s) + 3 h2o (â) â 2 nh3 (g) + 3 mgo (s) the yield of mgo is 3.60 g. what is the percent yield in the reaction?
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Ответ:
Mg₃N₂ (s) + 3 H₂O (l) --> 2 NH₃ (g) + 3 MgO (s)
Molar mass of Mg₃N₂ = 100.95 g/mol
Molar mass of H₂O = 18 g/mol
Molar mass of MgO = 40.3 g/mol
Moles Mg₃N₂: 3.82/100.95 = 0.0378
Moles H₂O: 7.73/18 = 0.429
Theo H₂O required for available Mg₃N₂: 0.0378*3/1 = 0.1134 mol
Hence, the limiting reactant is Mg₃N₂.
Thus,
Theoretical Yield = 0.0378 mol Mg₃N₂ * 3 mol MgO/Mg₃N₂ * 40.3 g/mol
Theo Yield = 4.57 g
Percent Yield = Actual Yield/Theo Yield * 100
Percent Yield = 3.60 g/4.57 g * 100 = 78.77%
Ответ:
Explanation:
Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.
The electrons are filled according to Afbau's rule in order of increasing energies and thus the electronic configuration for bromine with atomic number of 35 and thus containing 35 electrons is:
The nearest noble gas is Argon with 18 electrons and thus electronic configuration for bromine in terms of noble gas is: