inecomVODO Streaming Movies CExpert Q87A Jump 2 Sapling Lear

ine.comVODO Streaming Movies CExpert Q87A Jump 2 Sapling Learning A 25.598 g sample of aqueous waste diluted with 79.889 g of water. A 10.762 g aliquot of this solution is then titrated with 0.1062 M HCI. It required 32.84 mL of the HCI solution to reach the methyl red endpoint. Calculate the weight percent NHs in the aqueous waste a fertilizer manufacturer contains ammonia. The sample is Number

Solution

Moles of HCl used = 32.84 x 0.1062 / 1000 = 0.003487 Moles

Moles of NH3 present in 10.762 gm = 0.003487 Moles  

Mass of NH3 present in 10.762 gm = 0.003487 x 17.03 = 0.05939 gm

Mass of NH3 present in 79.889 gm = 0.05939 x 79.889 / 10.762 = 0.44089 gm

Weight percentage of ammonia = 0.44089 x 100 / 25.598 = 1.72 %

Balanced equation: NH3 + HCl = NH4Cl Reaction type: synthesis