posted 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?
-All reactants are invested in stoichiometric amounts
- Sodium nitrate
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.
Thank you so much for your respond! How do I convert the moles to 246TNA?