<p>I have a question:</p>
<p>How do you find the final moles and the final molarity of species in a reaction? (especially when one of the elements oxidizes)</p>
<p>Thanks</p>
<p>I took it last year and I thought it was tough but liked it, I would have a better shot of explaining to you if you actually wrote out the actual problem</p>
<p>use M=mol/L to find mols of the first reactant. Then use stoichiometry to find the moles of the second reactant. Then use M=mol/L again to find the molarity of the second reactant. </p>
<p>You ARE talking about titrations right?</p>
<p>Here it is:</p>
<p>8 H2SO4 + 10 FeS04 + 2 KMnO4 -> 2 MnSO4 + K2SO4 + 8 H2O + 5 Fe2(SO4)3</p>
<p>The following reactants are mixed to form an initial 200 mL solution</p>
<p>20.0 mL 3.00 M H2SO4
15.0 g FeSO4
100.0 mL .250 M KMnO4
80.0 mL H20</p>
<p>I need help finding the final moles and molarity of each of the following species:</p>
<p>H (charge = +1)
SO4 (charge = -2)
Fe (charge = +2)
K (charge = +1)
MnO4 (charge = -1)
Mn (charge = +2)
Fe (charge = +3)</p>
<p>look up stoichiometry</p>
<p>that’s a pretty tough question. But all you need to do is find the Limiting Reagent first. Then apply the steps i listed above, and it should work.</p>
<p>Convert everything that you have in L or g into moles, and then use the equation to see how much products form</p>
<p>The limiting reagent is 0.06 mol H2SO4.</p>
<p>Now what do I do?</p>
<p>now convert to moles of ions. since there are 2 mols of H for every mole of SO4, the conversion factor is 1molH2So4/2molH ions</p>
<p>do that for each thing that you need to find, and then use molarity equation for concentration.</p>
<p>Yeah, I understand that…</p>
<p>But what about calculating the final moles for Fe (charge = +2) and Fe (charge = +3)? Do I just do the same thing?</p>
<p>yeah… it’s exactly the same. just treat them as separate ions, because they are. For FeSO4, it’s 1mol/1mol. for Fe2SO43, it’s 3molFe/2molSO4</p>
<p>Okay! I don’t need help anymore.
Thank you.</p>
<p>your welcome.</p>