24.711 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. the sample is diluted with 79.741 g of water. a 11.169 g aliquot of this solution is then titrated with 0.1045 m hcl . it required 28.42 ml of the hcl solution to reach the methyl red endpoint. calculate the weight percent nh3 in the aqueous waste.

The weight percent of NH3 in the aqueous waste is 1.40 %

Explanation:

Step 1: Data given

Mass of the sample = 24.711 grams

Mass of water = 79.741 grams

A 11.169 g aliquot of this solution is then titrated with 0.1045 M HCl .

It required 28.42 mL of the HCl solution to reach the methyl red endpoint

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

HCl + NH3 → NH4Cl

Step 3: Calculate moles of HCl

Moles HCl = Molarity HCl* volume

Moles HCl = 0.1045 M * 0.02842 L

Moles HCl = 0.00297 moles

Step 4: Calculate moles NH3

1 mole of HCl consumed, need 1 mole of NH3 to produce, 1 mole of NH4Cl

For 0.00297 moles of HCl ,we need 0.00297 moles NH3

There is 0.00297 mole of NH3 in 11.169  grams of solution

Moles NH3 = 0.00297/11.169 * 79.741 grams

Moles NH3 = 0.02026 moles

Step 5: Calculate mass of NH3

Mass of NH3 = moles NH3 * Molar mass NH3

Mass of NH3 =0.02026 moles * 17.03 g/mol

Mass of NH3 = 0.345 grams

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

weight percent = (0.345 grams / 24.711 grams)*100%

Weight percent = 1.40 %

The weight percent of NH3 in the aqueous waste is 1.40 %