Federal regulations set an upper limit of 50 parts per million
(p...

Question

Federal regulations set an upper limit of 50 parts per million
(ppm) of NH3 in the air in a work environment [that is, 50
molecules of NH31g2 for every million molecules in the air].
Air from a manufacturing operation was drawn through a
solution containing 1.00 * 102 mL of 0.0105 M HCl. The
NH3 reacts with HCl according to:
NH31aq2 + HCl1aq2¡NH4Cl1aq2
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. (b) How many ppm of NH3 were in the air?
(Air has a density of 1.20 g/L and an average molar mass of
29.0 g>mol under the conditions of the experiment.)

Accepted Answer

Hi everyone here we have a question telling us that who guidelines recommends that the amount of residual chlorine and drinking water should not be higher than five parts per million. The amount of chlorine can be determined by black hydration with potassium iodide. And it gives us the equation chlorine Aquarius plus two iodine. Aquarius forms to chloride Aquarius plus iodine Aquarius. The iodine formed as a result of the reaction above, is then tie traded with a solution of sodium theo sulfate, two sodium theO self Aquarius plus iodine. Aquarius results into sodium iodide Aquarius plus sodium touch with DNA. A 385.0 millimeter water sample was added with excess solid sodium iodide and the resulting solution was then tie traded against a 1.8 times 10 to the negative third molar sodium theo sulfate solution. The solution required 12.7 mm of sodium theo sulfate to reach the equivalent of calculate the concentration of chlorine in the water sample in parts per million units. So our first step here is going to be to calculate our milligrams of chlorine. So we have 12 0. mill leaders of sodium theo sulfate and we're going to multiply that by 10 to the negative third. Leaders over one millimeter we're going to multiply that By 1.8, 8 Times 10 to the negative 3rd moles of sodium theo sulfate over one liter because our polarity is moles over. Leaders and we're going to multiply that by one mole of iodine over two moles of sodium theo sulfate And we're going to multiply that by one mole of chlorine over one more of iodine. And we're going to multiply that by Our molar mass of chlorine, which is 70 .90 g over one more of chlorine. And we're going to multiply that by one mg of chlorine Over 10 to the negative 3rd grams of chlorine. So our milliliters of sodium theo sulfate are canceling out. Our leaders are canceling out. Our moles of sodium theo sulfate are canceling out are moles of iodine are canceling out. Our moles of coin are canceling out and our grams of chlorine are canceling out, leaving us with zero point 846 mg chlorine. Now we have to calculate the parts per million of chlorine, So that's going to equal 0. milligrams of chlorine divided by 85 zero Mill leaders, times one millimeter Over 10 to the negative 3rd, leaders and our middle leaders are going to cancel out here And that gives us 2.20 parts per million. And that is our final answer. Thank you for watching. Bye

Watch next