Taking into account the reaction stoichiometry, the mass of silver(I) chloride formed is 930.37 grams.
In first place, the balanced reaction is:
2 AgNO₃ + BaCl₂ → 2 AgCl + Ba(NO₃)₂
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:
AgNO₃: 2 moles ×169.87 g/mole= 339.74 grams
BaCl₂: 1 mole ×208.24 g/mole= 208.24 grams
AgCl: 2 moles ×143.32 g/mole= 286.64 grams
Ba(NO₃)₂: 1 mole×261.34 g/mole= 261.34 grams
The following rule of three can be applied: if by reaction stoichiometry 208.24 grams of barium chloride form 286.64 grams of silver(I) chloride, 675.9 grams of barium chloride form how much mass of silver(I) chloride?
[tex]mass of silver(I) chloride=\frac{675.9 grams of barium chloride x286.64 grams of silver(I) chloride }{208.24 grams of barium chloride}[/tex]
mass of silver(I) chloride= 930.37 grams
Finally, the mass of silver(I) chloride formed is 930.37 grams.
Learn more about the reaction stoichiometry:
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