Respuesta :
the balanced equation for the above reaction is as follows;
2K + Cl₂ --> 2KCl
stoichiometry of K to KCl is 2:2
the number of KCl moles produced - 156 g / 74.5 g/mol = 2.09 mol
we have been given that Cl₂ is in excess. therefore K is the limiting reactant.
amount of product formed depends on number of limiting reactant present,
2 mol of K reacts to give 2 mol of KCl
therefore 2.09 mol of KCl being formed means - 2.09 mol of K has reacted
mass of K reacted is - 2.09 mol x 39 g/mol = 81.51 g of K
mass of K reacted is 81.51 g
2K + Cl₂ --> 2KCl
stoichiometry of K to KCl is 2:2
the number of KCl moles produced - 156 g / 74.5 g/mol = 2.09 mol
we have been given that Cl₂ is in excess. therefore K is the limiting reactant.
amount of product formed depends on number of limiting reactant present,
2 mol of K reacts to give 2 mol of KCl
therefore 2.09 mol of KCl being formed means - 2.09 mol of K has reacted
mass of K reacted is - 2.09 mol x 39 g/mol = 81.51 g of K
mass of K reacted is 81.51 g
The mass of Potassium required has been 81.79 g.
The chemical equation for the formation of Potassium chloride has been:
[tex]\rm 2\;K\;+\;Cl_2\;\rightarrow\;2\;KCl[/tex]
Computation for the mass of Potassium chloride
From the balanced chemical equation, for the formation of 2 moles of potassium chloride, 2 moles of potassium has been required.
Thus, for 1 mole Potassium chloride, 1 mole of potassium has been required.
The mass of 1 mole of compound has been equivalent to the molar mass of the compound.
The molar mass of potassium has been 39.09 grams, and the molar mass of potassium chloride has been 74.55 grams.
Thus, the amount of potassium required has been given as:
[tex]\rm 74.55\;g\;KCl=39.09\;g\;K\\\\
156\;g\;KCl=\dfrac{39.09}{74.55}\;\times\;156\;g\;K\\\\
156\;g\;KCl=81.79\;g\;K[/tex]
Thus, the mass of Potassium required has been 81.79 g.
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