Considering the reaction stoichiometry and the definition of limiting reagent, the mass of NaNO₃ that is produced when 25.0 grams of lead (II) nitrate react with 15.0 grams of sodium iodide is 8.50 grams.
The balanced reaction is:
Pb(NO₃)₂ + 2 Nal → PbI₂ + 2 NaNO₃
Reaction stoichiometry
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:
- Pb(NO₃)₂: 1 mole
- Nal: 2 moles
- PbI₂: 1 mole
- NaNO₃: 2 moles
The molar mass of the compounds present in the reaction is:
- Pb(NO₃)₂: 331 g/mole
- Nal: 149.9 g/mole
- PbI₂: 460.8 g/mole
- NaNO₃: 85 g/mole
Then, by reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of mass of each compound participate in the reaction:
- Pb(NO₃)₂: 1 mole× 331 g/mole= 331 grams
- Nal: 2 moles× 149.9 g/mole= 299.8 grams
- PbI₂: 1 mole× 460.8 g/mole= 460.8 grams
- NaNO₃: 2 moles× 85 g/mole= 170 grams
Limiting reagent
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 331 grams of Pb(NO₃)₂ reacts with 299.8 grams of NaI, 25 grams of Pb(NO₃)₂ reacts with how much mass of NaI?
[tex]mass of NaI=\frac{25 grams of Pb(NO_{3} )_{2} x299.8 grams of NaI}{331 grams of Pb(NO_{3} )_{2} }[/tex]
mass of NaI= 22.64 grams
But 22.64 grams of NaI are not available, 15 grams are available. Since you have less mass than you need to react with 25 grams of Pb(NO₃)₂, NaI will be the limiting reagent.
Mass of NaNO₃ formed
Then, it is possible to determine the mass of NaNO₃ produced by another rule of three, taking into account the limiting reagent: if by stoichiometry 299.8 grams of NaI produce 170 grams of NaNO₃, 15 grams of NaI how much mass of NaNO₃ will be formed?
[tex]mass of NaNO_{3} =\frac{15 grams of NaIx170 grams of NaNO_{3}}{299.8 grams of NaI}[/tex]
mass of NaNO₃= 8.50 grams
In summary, the mass of NaNO₃ that is produced when 25.0 grams of lead (II) nitrate react with 15.0 grams of sodium iodide is 8.50 grams.
Learn more about reaction stoichiometry:
brainly.com/question/11999193
brainly.com/question/15111313?referrer=searchResults
brainly.com/question/6061451?referrer=searchResults
