Air from a manufacturing operation was drawn through a solution containing 1.00

Question

Air from a manufacturing operation was drawn through a solution containing 1.00 x 102 mL of 0.0105 M HCl. The NH3 reacts with HCl as follows:

NH3(aq) +HCl(aq) --> NH4Cl(aq)

After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point. How many grams of NH3 were drawn into the acid solution? How many ppm of NH3 were in the air assuming air has a density of 1.20 g/L and an average molar mass of 29.0 g/mol?

Very confused please help :( !!

Explanation / Answer

moles of HCl initially = 0.0105*0.1 = 0.00105 moles

moles of NaOH needed = 0.0588*0.0131 = 0.00077 moles

so mole of HCl that reacted with NH3 = 0.00105-0.00077 = 0.00028 moles of HCl = moles on NH3

so mass of NH3 = 0.00028*17 = 0.00476gm = 4.76 mg

total volume of air = 10*10 = 100 L

so mass of Air = 100*1.02 = 102 gm

so ppm = 0.00476/102 *1000000= 46.667 ppm

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