<li>Bacterial digestion is an economical method of sewage treatment. The reaction</li>
</ol>
<p>5CO2(g) + 55NH4+ (aq) + 76O2 (g) —> C5H7O2N (s) + 54NO2- (aq) + 52H2O(l) + 109H+ (aq)</p>
<p>is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions. What mass of bacterial tissue is produced in a treatment plant for every 1.0 x 10^2 kg of waste water containing 3.0% NH4+ ions by mass? Assume that 95% of the ammonium ions are consumed by the bacteria.</p>
<p>I’m confused: there are no nitrate ions here in this equation? NO2 is nitrite, and that’s the closest to it.</p>
<p>OK- this is what I think it’s saying. Because C5H7O2N is an organic compound and the other products are nitrous acid, water, and hydrogen, I believe that C5H7O2N is the “bacterial tissue” that is “produced” in this equation. </p>
<p>According to the information, there is 3.0 kg of NH4+ ions in the water. However, we must also consider that only 95% of the ions will be used in the equation, and thus there is 2.85 kg of NH4+ ions used (there are 2 significant figures, but you should never round it until the end, so we’ll keep it as 2.85).</p>
<p>We must now figure out the number of moles in NH4+ ions, so that we can figure out the # of moles of C5H7O2N later on. The molar mass of NH4+ is 14.007 + (4 x 1.0079) = 18.039 g/mol.</p>
<p>2.85 kg x (1000 g / 1 kg) x (1 mol / 18.039 g) = 157.991019 mol</p>
<p>Remember the coefficients form a molar ratio, so:</p>
<h1>moles NH4+</h1>
<p>--------------- = 55 / 1 = 55</p>
<h1>mole C5H702N</h1>
<p>The number of moles of C5H7O2N is 157.991019 / 55 = 2.87256398 mol.</p>
<p>Now, to find the mass of the tissue, we have to find the molar mass of C5H7O2N, which is (5 x 12.011) + (7 x 1.0079) + (2 x 16.00) + 14.007 = 113.12 g/mol.</p>
<p>Let’s find the mass.</p>
<p>2.87256398 mol x (113.12 g/ mol) = 324.944437 g</p>
<p>We originall were given kg as the units, so let’s use it</p>
<p>324.944437 g x (1 kg / 1000 g) = 0.324944437 kg </p>
<p>Remember there are 2 significant figures, so let’s round now.</p>
<p>The answer is 0.32 kg.</p>
<p>I don’t really understand how do do calculations involving limiting reactants. Can someone explain it better than Zumdahl’s textbook?</p>
<ol>
<li>Hydrogen peroxide is used as a cleaning agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes upon contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams upon contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide:</li>
</ol>
<p>BaO2 (s) + 2HCl (aq) —> H2O2 (aq) + BaCl2 (aq)</p>
<p>What mass of hydrogen peroxide should result when 1.50 g of barium peroxide is treated with 25.0 mL of hydrochloric acid solution containing 0.0272 g of HCl per mL. What mass of which reagent is left unreacted?</p>
<p>Ok, I know what you mean, let me explain the method that fits into my stupid head. So, the equation tells us that for every barium peroxide molecule there needs to be 2 hydrochloric acid molecules to result in 1 molecule of hydrogen peroxide. Therefore the proportion of BaO2 : HCl is 1:2. Now to see which molecule is the limiting reactant you need to see the real proportion of molecules at hand.
1.5 g BaO2 * (1 mol BaO2 / 169.3 g BaO2) = .00886 mol BaO2
25 mL HCl * (.0272g / mL) * (1 mol HCl / 36.45 g HCl) = .01867 mol HCl</p>
<p>Notice that the proportion of HCl to BaO2 in this case is 2.1 molecules HCl: 1 molecule BaO2.
Because this exeeds the 2:1 ration we established, HCl is in excess and BaO2 is the limiting reactant.</p>
<p>To find the mass of the H2O2, use the # moles of limiting reactant, because only that amount will actually react. In case of HCl only .01867 moles will react and 9.5 *10^-4 moles of HCl will be left unreacted.
So,
.00886 mol BaO2 * (1 mol H2O2 / 1 mol BaO2) * (34 g H2O2/ 1 mol H2O2) = .301 g H2O2.
Hope that helps, it helps me.</p>
<p>Lol our class used Zumdahl’s textbook. I never listened in class ever but read the Zumdahl assigned readings… got a 5. Read that dam thing.</p>
<p>So basically we have to find all the possible proportions and check?</p>
<p>To find the mass of the unreacted reagent, do I do this?</p>
<p>2.1 mol HCl/1 mol BaO2. The original ratio was 2:1, so .1 of HCl is unreacted? Does that make sense?</p>
<p>.1 x 0.1867 mol HCl x (36.46g HCl/ 1 mol HCl) = .681g HCl left unreacted.</p>
<p>Is this right?</p>
<p>no… it just means that you have .1 mol HCl excess for every mol BaO2 you have. Because you have 0.00886 mol BaO2 according to ktoto (im not in the mood to do the math to check) then it’s really
.1 mol excess HCl / 1 mol BaO2 = x mol excess HCl / 0.00886 mol BaO2</p>
<p>Solving for the proportion would hopefully get .301 / 34, which is the mass of the H202 divided molar mass of H2O2 (getting the # of moles).</p>
<p>It’s much easier to thinking along ktoto’s lines and it’s less prone to error.</p>
<p>Another way of thinking of it is: know the correct proportion using the coefficients (the “theoretical” proportion). Observe the proportion of the stuff that you’re given (the “actual” proportion). If the actual proportion is less in value than the theoretical one, the compound used for the numerator of the proportion is the limiting reactant. If the actual proportion is more, the compound used for the denominator of the proportion is the limiting reactant.</p>
<p>No, dont over-interpret the actual ratio, to find out the amount of unreacted HCl do this:
Knowing that BaO2 is the limiting reactant you know that its completely going to be consumed.
.00886 mol BaO2 will be consumed and because of the 1:2 ration from the cooficients only
.00886 mol BaO2 * ( 2 mol HCl / 1 mol BaO2) = .01772 mol HCl will be consumed
We had .01867 mol HCl, only .01772 mol was consumed, so we have left
.00095 mol HCl, which is about .0346 g of HCl.</p>