The explosive residue contains carbon, hydrogen and nitrogen where the percentage of nitrogen is high, calculate it from 0.5g of residue which eliminates ammonia gas on reaction with sodium hydroxide, and absorb by 50mL of Nby10 H2SO4. Unused acid is neutralized/titrated with 14mL potassium hydroxide
i am using the formula
% of nitrogen = (1.4*Nacid*Vacid)/(weight of residue)
is the answer 14%
or 3.9%
I can't tell what you've done.
To calculate the percentage of nitrogen in the explosive residue, we can use the given information and the formula you mentioned.
The formula you provided is:
% of nitrogen = (1.4 * Nacid * Vacid) / (weight of residue)
Let's break down the given information and use it in the formula:
1. We have 0.5g of residue.
2. The residue eliminates ammonia gas on reaction with sodium hydroxide, which means that the remaining nitrogen in the residue is determined by the ammonia reacted with the sodium hydroxide. So, the nitrogen present in the residue is the nitrogen in ammonia.
3. The ammonia gas is absorbed by 50mL of Nby10 H2SO4. This means that the reaction with sodium hydroxide neutralizes the ammonia gas using 50mL of H2SO4.
4. The unused H2SO4 is neutralized/titrated with 14mL potassium hydroxide. This step allows us to quantify the amount of H2SO4 remaining after the reaction.
Now, let's put these values into the formula:
% of nitrogen = (1.4 * Nacid * Vacid) / (weight of residue)
Assuming Nacid is the normality of H2SO4 and Vacid is the volume of H2SO4 used for titration:
Nacid = 10 (since it's Nby10 H2SO4)
Vacid = 14 mL (as it neutralizes/titrates with 14mL of potassium hydroxide)
Weight of residue = 0.5g
Substituting the values, we get:
% of nitrogen = (1.4 * 10 * 14) / 0.5
Calculating this expression, we find that the answer is approximately 78.4%.
Therefore, the percentage of nitrogen in the explosive residue is 78.4%, not 14% or 3.9%.