A 24.393 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 80.601 g of water. A 12.404 g aliquot of this solution is then titrated with 0.1059 M HCl. It required 31.61 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

To calculate the weight percent (wt%) of NH3 in the aqueous waste, we need to determine the number of moles of NH3 in the 12.404 g aliquot, and then use that value to find the wt% NH3 in the original sample.

Step 1: Calculate the number of moles of HCl used in the titration.

Using the equation n = C x V, where n represents the number of moles, C is the concentration, and V is the volume in liters, we can calculate the moles of HCl:

n(HCl) = C(HCl) x V(HCl)
n(HCl) = (0.1059 mol/L) x (31.61 mL / 1000 mL/L)
n(HCl) = 0.00334 moles

Step 2: Write the balanced chemical equation for the reaction between NH3 and HCl.

NH3 + HCl -> NH4Cl

Step 3: Determine the moles of NH3 reacted based on the moles of HCl used.

From the balanced equation, we can see that the ratio of NH3 to HCl is 1:1. Therefore, the number of moles of NH3 is also 0.00334 moles.

Step 4: Calculate the moles of NH3 in the 12.404 g aliquot.

Using the molar mass of NH3 (17.031 g/mol), we can calculate the moles of NH3 in the 12.404 g aliquot:

moles of NH3 = mass / molar mass
moles of NH3 = 12.404 g / 17.031 g/mol
moles of NH3 = 0.7271 moles

Step 5: Calculate the moles of NH3 in the original sample.

Since the 12.404 g aliquot was diluted with 80.601 g of water, the mass of the original sample is the sum of the two:

mass of original sample = mass of aliquot + mass of water
mass of original sample = 12.404 g + 80.601 g
mass of original sample = 92.005 g

Using the moles of NH3 in the aliquot and the mass of the original sample, we can find the moles of NH3 in the original sample:

moles of NH3 in original sample = (moles of NH3 in aliquot / mass of aliquot) x mass of original sample
moles of NH3 in original sample = (0.7271 moles / 12.404 g) x 92.005 g
moles of NH3 in original sample = 5.3855 moles

Step 6: Calculate the weight percent NH3 in the aqueous waste.

weight percent NH3 = (moles of NH3 / mass of sample) x 100
weight percent NH3 = (5.3855 moles / 92.005 g) x 100
weight percent NH3 = 5.849%

Therefore, the weight percent of NH3 in the aqueous waste is approximately 5.849%.

To calculate the weight percent of NH3 in the aqueous waste, we need to determine the amount of NH3 in the 12.404 g aliquot of the solution and then convert it to a percentage.

First, let's calculate the number of moles of HCl used in the titration. We can use the concentration and volume of the HCl solution:

moles HCl = concentration × volume
moles HCl = 0.1059 M × 0.03161 L
moles HCl = 0.0033465 mol

Since NH3 is a weak base, it reacts with HCl in a 1:1 ratio:

moles NH3 = 0.0033465 mol

The molar mass of NH3 is 17.03 g/mol:

mass NH3 = moles NH3 × molar mass
mass NH3 = 0.0033465 mol × 17.03 g/mol
mass NH3 = 0.05694 g

Now, let's calculate the weight percent of NH3 in the aqueous waste:

weight percent NH3 = (mass NH3 / mass of solution) × 100%
weight percent NH3 = (0.05694 g / 24.393 g) × 100%
weight percent NH3 = 0.2338%

Therefore, the weight percent of NH3 in the aqueous waste is approximately 0.2338%.