Asked by jocckee
2,4,6-Trinitroanilin can be prepared by nitration of p-nitroaniline with nitrate ions in the acidic solution. Chemical engineer sounds 0.50 grams of p-nitroaniline (dissolved in sulfuric acid) to react with 0.60 g of sodium nitrate (dissolved in sulfuric acid). Which reactant is limiting?
options:
-All reactants are invested in stoichiometric amounts
- 2,4,6-Trinitroanilin
- P-Nitroaniline
- Water
- Sodium nitrate
options:
-All reactants are invested in stoichiometric amounts
- 2,4,6-Trinitroanilin
- P-Nitroaniline
- Water
- Sodium nitrate
Answers
Answered by
DrBob222
p-nA + NaNO3 ==> 246TNA
mols p-nitroaniline = grams/molar mass
mols NaNO3 = grams/molar mass
Using the coefficients in the balanced equation, convert mols p-nA to 246TNA.
Do the same for mols NaNO3 to mols 246TNA
It is likely the two values will be different; in limiting reagent problems the correct value is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
mols p-nitroaniline = grams/molar mass
mols NaNO3 = grams/molar mass
Using the coefficients in the balanced equation, convert mols p-nA to 246TNA.
Do the same for mols NaNO3 to mols 246TNA
It is likely the two values will be different; in limiting reagent problems the correct value is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Answered by
jocckee
Thank you so much for your respond! How do I convert the moles to 246TNA?
:-)
:-)
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