10. NEED HELP FAST

Question

A  20.574  g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with  71.827  g of water. A  10.093  g aliquot of this solution is then titrated with  0.1062  M  HCl . It required  30.09  mL of the  HCl  solution to reach the methyl red endpoint.

Calculate the weight percent  NH3  in the aqueous waste.

wt%  NH3=

Answer 1

Fun question!

You start with the titration and work backwards.

30.09 mL of 0.1062M HCL = (0.03009 L) * 0.1062 mol/L = 0.003196 mol HCL = 0.003916 mol NH3

0.003916 mol NH3 * 17.031 g/mol = 0.05442 g NH3 in the 10.093 g aliquot

0.05442 g/10.093 g = 0.5392% m/m

20.574 g original H2O-NH3 + 71.827 g H2O = 92.401 g H2O-NH3

0.005392 * 92.401 g = 0.4982 g NH3

0.4982 g NH3/20.574 g = 2.42% NH3