In the reaction of 675.9 grams of barium chloride and excess silver(I) nitrate, how many grams of silver(I) chloride should get
1 answer:
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= 930.37 grams
Finally, the mass of silver(I) chloride formed is 930.37 grams.
Learn more about the reaction stoichiometry:
#SPJ1
You might be interested in
Aueyeowosy yo this is my answer
