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Cloud [144]
1 year ago
5

In the reaction of 675.9 grams of barium chloride and excess silver(I) nitrate, how many grams of silver(I) chloride should get

formed?
i dont understand, show work pls ty
Chemistry
1 answer:
zvonat [6]1 year ago
8 0

Taking into account the reaction stoichiometry, the mass of silver(I) chloride formed is 930.37 grams.

<h3>Reaction stoichiometry</h3>

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:

  • AgNO₃: 2 moles
  • BaCl₂: 1 mole
  • AgCl: 2 moles
  • Ba(NO₃)₂: 1 mole

The molar mass of the compounds is:

  • AgNO₃: 169.87 g/mole
  • BaCl₂: 208.24 g/mole
  • AgCl: 143.32 g/mole
  • Ba(NO₃)₂: 261.34 g/mole

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

<h3>Mass of silver(I) chloride formed</h3>

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?

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}

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:

brainly.com/question/24741074

brainly.com/question/24653699

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