1. Air from a manufacturing operation was drawn through a solution containing 1.
ID: 812932 • Letter: 1
Question
1. Air from a manufacturing operation was drawn through a solution containing 1.00 x 102 ml of a 0.0405 M hydrobromic acid. After drawing air through the acid solution for 10.0 minutes at a rate of 10.0 L/minute the acid was titrated. To neutralize the acid it took 26.5mL of a 0.0588 M sodium hydroxide.
A) write a balanced chemical equation including all phases for the neutralization reaction from above.
B) how much excess acid remains from the above reaction?
To measure the amount of ammonia in the air the remaining excess acid reacts with ammonia to form ammonium bromide shown in the equation below :
NH3(aq) + HBr(aq) --> NH4Br
C) how many grams of ammonia were used up in the reaction?
D) how many ppm of ammonia were in the air? Given that density of air is 1.20 g/L and the average molar mass of air is 29.0 g/mol under the conditions in this experiment.
E) federal regulations set an upper limit of 50 ppm of NH3, in the air in a work environment. Is this manufacturer in compliance with federal regulations?
Explanation / Answer
A) HBr(aq) + NaOH(aq) ---------> NaBr(aq) + H2O(l)
B) Mole of HBr in 100mL of 0.0405M = 0.00405 moles
Mole of NaOH added = 0.0588*0.0265 = 0.00156
Therefore, the excess acid is = 0.00405 - 0.00156 = 0.00249 moles
C) 0.00249 moles of HBr would react with same amount of NH3. Therefore, mass of NH3 is 0.00249*17 = 0.0424g
D) density of air is 1.2 g/L. Total amount of air passed through HBr solution is 10*10 = 100 L and the corresponding mass = 1.2*100 = 120g or 120/29 = 4.138 mol.
Therefore the mole fraction of NH3 in air = 0.00249/4.138 = 0.01 or 10000 ppm by moles.
ppm by mass = 0.0424/120 = 353.3 ppm
E) The level of NH3 in air exceeds the upper limit set by federal regulations. Therefore this manufacturer is not in compliance with federal regulations.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.