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Federal regulations set an upper limit of 50 ppm of NH3 in the air in a work environment. Air from a manufacturing operation was drawn through a solution containing 100 mL of .0105 M HCl. The NH3 reacts with HCl:
NH3(aq) + HCl(aq)--> NH4Cl(aq)

After drawing air through the acid solution for 10.0 min. at a rate of 10.0L/min, the acid was titrated. The remaining acid needed 13.1 mL of .0588 M NaOH to reach the equivalence point.
a) How many grams of NH3 were drawn into the acid solution?
b) How many ppm of NH3 were in the air (using air density= 1.2 g/L and an average molar mass of 29.0 g/mol)?

2007-05-31 11:32:32 · 1 answers · asked by Anonymous in Science & Mathematics Chemistry

1 answers

a) 4.76 mg

b) 39.6 ppm

2007-06-01 05:53:30 · answer #1 · answered by ag_iitkgp 7 · 0 0

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