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

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?

:-)

To determine which reactant is limiting, we need to compare the number of moles of each reactant and their stoichiometric ratios.

First, we need to calculate the number of moles of each reactant:

Molar mass of p-nitroaniline (C6H6N2O2):
C = 12.01 g/mol × 6 = 72.06 g/mol
H = 1.008 g/mol × 6 = 6.048 g/mol
N = 14.01 g/mol × 2 = 28.02 g/mol
O = 16.00 g/mol × 2 = 32.00 g/mol

Total molar mass = 72.06 g/mol + 6.048 g/mol + 28.02 g/mol + 32.00 g/mol = 138.13 g/mol

Number of moles of p-nitroaniline = mass / molar mass
Number of moles of p-nitroaniline = 0.50 g / 138.13 g/mol

Now, let's calculate the number of moles of sodium nitrate:

Molar mass of sodium nitrate (NaNO3):
Na = 22.99 g/mol
N = 14.01 g/mol
O = 16.00 g/mol × 3 = 48.00 g/mol

Total molar mass = 22.99 g/mol + 14.01 g/mol + 48.00 g/mol = 85.00 g/mol

Number of moles of sodium nitrate = mass / molar mass
Number of moles of sodium nitrate = 0.60 g / 85.00 g/mol

Now, let's calculate the stoichiometric ratio between p-nitroaniline and 2,4,6-Trinitroanilin:

From the balanced chemical equation, we know that the ratio between p-nitroaniline and 2,4,6-Trinitroanilin is 1:1.

Now we can compare the number of moles of each reactant:

Number of moles of p-nitroaniline = 0.50 g / 138.13 g/mol = x mol
Number of moles of sodium nitrate = 0.60 g / 85.00 g/mol = y mol

Since the stoichiometric ratio is 1:1 between p-nitroaniline and 2,4,6-Trinitroanilin, the number of moles of p-nitroaniline is equal to the number of moles of 2,4,6-Trinitroanilin.

Therefore, the reactant that is limiting is p-nitroaniline.