Answer:
0.322630 mols
Explanation:
[tex]\frac{21.0g Zn solid}{65.09 Zn}[/tex]
You have to divide the given by the molar mass then multiply by the mol ratio which in this case is 1:1
Taking into account the reaction stoichiometry, 0.32 moles of ZnCl₂ are formed when 21 grams of Zn reacts assuming CuCl₂ is available in excess.
In first place, the balanced reaction is:
Zn + CuCl₂ → ZnCl₂ + Cu
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The molar mass of the compounds is:
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
Zn: 1 mole ×65.37 g/mole= 65.37 grams
CuCl₂: 1 mole ×134.44 g/mole= 134.44 grams
ZnCl₂: 1 mole ×136.27 g/mole= 136.27 grams
Cu: 1 mole ×63.54 g/mole= 63.54 grams
The following rule of three can be applied: if by reaction stoichiometry 65.37 grams of Zn form 1 mole of ZnCl₂, 21 grams of Zn form how many moles of ZnCl₂?
[tex]moles of ZnCl_{2} =\frac{21 grams of Znx1 mole of ZnCl_{2} }{65.37 grams of Zn}[/tex]
moles of ZnCl₂= 0.32 moles
Then, 0.32 moles of ZnCl₂ are formed when 21 grams of Zn reacts assuming CuCl₂ is available in excess.
Learn more about the reaction stoichiometry:
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