Question

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.

Answer 1

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

Step-by-step 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 %