This is how I would go about solving this problem.
moles of NH3 in sample= 0.1072M * 29.32 x 10^-3L=3.14 x 10^-3 moles of NH3.
Remember, the mole ratios are equal since it takes one mole of NH3 to react with one mole of HCl.
3.14 x 10^-3 moles of NH3*(17.031g NH3/ 1mole)= mass of NH3
(mass of NH3/mass of aliquot sample)= ratio of NH3 in sample after it was diluted
ratio of NH3 in sample after it was diluted* total weight after dilution= total mass of of NH3 in sample
(total mass of of NH3 in sample/22.617 g sample of aqueous waste)*100= wt% NH3
*****I left a note for you in one of your other posts about a correction.
A 22.617 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 70.407 g of water. A 10.883 g aliquot of this solution is then titrated with 0.1072 M HCl. It required 29.32 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.
wt% NH3 = ??
Hint: The titration equation shows one mole of HCl reacts with one mole of NH3:
HCl + NH3 <---> NH4^+ + Cl^-
Haven't covered this topic in class yet. How to solve??
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1 answer