How many grams of Ag2S2O3 form
1 answer:
Taking into account the reaction stoichiometry, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
2 AgBr + Na₂S₂O₃ → Ag₂S₂O₃ + 2 NaBr
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:
- AgBr: 2 moles
- Na₂S₂O₃: 1 mole
- Ag₂S₂O₃: 1 mole
- NaBr: 2 moles
The molar mass of the compounds is:
- AgBr: 187.77 g/mole
- Na₂S₂O₃: 158 g/mole
- Ag₂S₂O₃: 327.74 g/mole
- NaBr: 102.9 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- AgBr: 2 moles ×187.77 g/mole= 375.54 grams
- Na₂S₂O₃: 1 mole ×158 g/mole= 158 grams
- Ag₂S₂O₃: 1 mole ×327.74 g/mole= 327.74 grams
- NaBr: 2 moles ×102.9 g/mole= 205.8 grams
The following rule of three can be applied: if by reaction stoichiometry 375.54 grams of AgBr form 327.74 grams of Ag₂S₂O₃, 125 grams of AgBr form how much mass of Ag₂S₂O₃?
<u><em>mass of Ag₂S₂O₃= 109.09 grams</em></u>
Then, 109.09 grams of Ag₂S₂O₃ are formed when 125 g AgBr reacts completely.
Learn more about the reaction stoichiometry:
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