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. (a) How many grams of NH3 were drawn into the
acid solution?

Accepted Answer

Hi everyone here we have a question that says the who guidelines recommend that the amount of residual chlorine in drinking water should not be higher than five parts per million. The amount of corn can be determined by backed iteration with potassium iodide and that equation is chlorine nucleus 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 the solution of sodium theo sulfate to sodium theo sulfate plus iodine. Aquarius forms to sodium iodide Aquarius plus sodium teacher Thean eight A 350.0 mL. Water sample was added with excess solid sodium iodide and the resulting solution was then titrate ID against a 2.25 times 10 to the negative three molar sodium theo sulfate solution. The solution required 14.5 ml of sodium theo sulfate to reach the equivalence point. How many grams of chlorine are there in the water sample? So we're going to start off with our 14.5 ml of sodium Theosophy eight And we're going to multiply that by 10 to the -3. Leaders over one millimeter And we're going to multiply that by 2. Times 10 to the -3 moles of sodium theo sulfate over one leader because polarity is moles over leaders And we're going to multiply that by one mole of I diane over two moles of sodium theo sulfate And we're going to multiply that by one mole of chlorine over one mole of iodine? And we're going to multiply that by chlorine molar mass, which is 70 .90 g of chlorine Over one mole of coin. So our middle leaders of sodium theo sulfate are canceling out. Our leaders are canceling out are moles of sodium theo sulfate are canceling out. Our moles of iodine are canceling out and our moles of chloride are canceling out. And that gives us 0. 1157. If we want to put that into scientific notation, that would be 1.16 times 10 to the negative three grams of chlorine. And that is our final answer. Thank you for watching. Bye.

Watch next