A 23.596 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 79.504 g of water. A 13.803 g aliquot of this solution is then titrated with 0.1098 M HCl. It required 29.02 ml of the HCl solution to reac the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.
HCl + NH3 ---> NH4^+ +Cl^-
I have to find the moles of hcl consumed , convert them to moles of NH3 and then convert that to grams of NH3 but i really don't know how to do each one of these steps. Help?
moles HCl = M x L = 0.1098M x 0.02902L = approximately 0.003
The NH3 + HCl equation is balanced and it is 1 mole NH3 to 1 mole HCl; therefore, moles NH3 = moles HCl.
grams NH3 = moles NH3 x molar mass NH3 = approximately 0.05 g NH3 in the titrated sample (the 13.803 g portion). You need to convert this to grams in the sample. You do that by
0.05 g in the titration part x (23.596 + 79.504)/(13.803 = approximately 0.4 g in the original sample. Then
%NH3 = (grams NH3/mass sample)*100 = ??
mass of the sample is 23.596
I got 1.6134% which doesnt make sense :s
To calculate the weight percent of NH3 in the aqueous waste, we need to follow these steps:
Step 1: Find the moles of HCl consumed.
Step 2: Use the balanced equation to convert the moles of HCl to moles of NH3.
Step 3: Convert moles of NH3 to grams of NH3.
Step 4: Calculate the weight percent of NH3 in the aqueous waste.
Let's go through each step in detail:
Step 1: Find the moles of HCl consumed.
Given:
- Volume of HCl solution used = 29.02 mL = 0.02902 L (convert milliliters to liters)
- Concentration of HCl solution = 0.1098 M (moles per liter)
Moles of HCl consumed = Volume of HCl solution used × Concentration of HCl solution
= 0.02902 L × 0.1098 M
≈ 0.003191 moles (rounded to 4 decimal places)
Step 2: Use the balanced equation to convert the moles of HCl to moles of NH3.
Looking at the balanced equation: HCl + NH3 ---> NH4+ + Cl-
We can see that 1 mole of HCl reacts with 1 mole of NH3.
Therefore, moles of NH3 = moles of HCl = 0.003191 moles
Step 3: Convert moles of NH3 to grams of NH3.
To do this, we need the molar mass of NH3, which is:
Molar mass of NH3 = (1 × Molar mass of N) + (3 × Molar mass of H)
= (1 × 14.01 g/mol) + (3 × 1.01 g/mol)
= 14.01 g/mol + 3.03 g/mol
= 17.04 g/mol
So, grams of NH3 = moles of NH3 × Molar mass of NH3
= 0.003191 moles × 17.04 g/mol
≈ 0.05447 grams (rounded to 5 decimal places)
Step 4: Calculate the weight percent of NH3 in the aqueous waste.
The weight percent is calculated by dividing the mass of NH3 by the total mass of the solution (sample + water) and multiplying by 100.
Weight percent of NH3 = (grams of NH3 / total mass of the solution) × 100
Given:
- Mass of the sample = 23.596 g
- Mass of water added = 79.504 g
Total mass of the solution = Mass of the sample + Mass of water added
= 23.596 g + 79.504 g
= 103.1 g
Weight percent of NH3 = (0.05447 grams / 103.1 grams) × 100
≈ 0.0528% (rounded to 3 decimal places)
Therefore, the weight percent of NH3 in the aqueous waste is approximately 0.0528%.
To solve this question, we need to go step by step:
Step 1: Calculate the moles of HCl used in the titration.
Given that 29.02 mL of 0.1098 M HCl was required, we can calculate the moles of HCl as follows:
Moles of HCl = concentration of HCl (in M) * volume of HCl used (in L)
First, convert the volume of HCl used from mL to L:
29.02 mL = 29.02 mL * (1 L / 1000 mL) = 0.02902 L
Now, calculate the moles of HCl:
Moles of HCl = 0.1098 M * 0.02902 L = 0.003182 Moles
Step 2: Determine the stoichiometry between HCl and NH3 using the balanced equation:
HCl + NH3 ---> NH4+ + Cl-
From the balanced equation, we can see that one molecule of HCl reacts with one molecule of NH3. Therefore, the moles of HCl consumed will be in the same ratio as the moles of NH3 produced.
Step 3: Calculate the moles of NH3.
Since the moles of HCl are equal to the moles of NH3, we have:
Moles of NH3 = 0.003182 Moles
Step 4: Convert the moles of NH3 to grams of NH3.
To convert the moles of NH3 to grams, we need the molar mass of NH3.
The molar mass of NH3 is calculated as follows:
Molar mass of NH3 = (1 * molar mass of N) + (3 * molar mass of H)
= (1 * 14.01 g/mol) + (3 * 1.01 g/mol)
= 14.01 g/mol + 3.03 g/mol
= 17.04 g/mol
Now, calculate the mass of NH3:
Mass of NH3 = Moles of NH3 * Molar mass of NH3
= 0.003182 Moles * 17.04 g/mol
= 0.05435 grams
Step 5: Calculate the weight percent of NH3 in the aqueous waste.
To determine the weight percent, we need to compare the mass of NH3 to the total mass of the solution.
Total mass of the solution = mass of the sample + mass of water
= 23.596 g + 79.504 g
= 103.1 g
Finally, calculate the weight percent of NH3:
Weight percent of NH3 = (Mass of NH3 / Total mass of the solution) * 100
= (0.05435 g / 103.1 g) * 100
= 0.05278 %
Therefore, the weight percent of NH3 in the aqueous waste is approximately 0.05278%.