<p>Hi, my little sister is having a test tommorow and she was having trouble with three problems. Ugh, I haven’t taken Chem since sophmore year, so I can’t really help her much. She needs to know how to solve these for her test tommorow. </p>
<li><p>Balance the following net ionic redox reaction in acid solution, then in basic solution using the half-reaction method.
Br^-1 + MnO4^-1 → Br2 + Mn^2+</p></li>
<li><p>68.0 g of sodium react with 60.0 g of water to produce sodium hydroxide and hydrogen gas, how many grams of each reactant and product reamin after reaction?</p></li>
<li><p>23.5 mL of a 0.0100 M Ca(OH)2 react with 15.0 ml of HCL, what is the molarity of HCL? HOw many grams of HCL theorhetically exist?</p></li>
</ol>
<p>Thanks.</p>
<p>1) is too much trouble to answer in this kind of setting... very strange that she's getting these kinds of questions in September - this can't h.s. chemistry. You have to determine oxidation numbers then find the electron transfers, balance the half reactions by charge then combine so that the electrons cancel. Add equations together and in an acid solution you add H+ to balance the charge and OH- is added in a basic solution. Balance hydrogen by adding H2O.</p>
<ol>
<li> 2Na + 2H2O --->2NaOH + H2</li>
</ol>
<p>68 g Na * 1 mole Na/23 g * 2 moles NaOH/2 moles Na --> 2.96 moles NaOH</p>
<p>60 g H2O * 1 mole H2O/18g * 2 moles NaOH/2 moles H2O ---> 3.33 moles NaOH</p>
<p>So Na is the limiting reagent and there will be
2.96 moles NaOH * 40 g NaOH/1mole NaOH --> 118.4 grams NaOH</p>
<p>2.96 moles NaOH * 1mole H2/2 moles NaOH * 2 g H2/1moleH2 ---> 2.96 grams H2</p>
<p>There will be no sodium left since it was the limiting reagent.<br>
2.96 moles NaOH * 2 moles H2O/ 2 moles NaOH * 18 g H2O/1mole H2O -->53.28 gram H2O used so.... you have 60-53.28 or 6.72 g H2O left after the reaction</p>
<p>Yes, she's taking AP Chemistry. (She's a Junior.) Thanks for the help.</p>
<p>3> Molarity * Volume of the Calcium Hydroxide = Molarity * Volume of HCl which will allow you to calculate the moles of HCL and then the grams of HCL. Just remember Ca+2 ion and H+1 ion.</p>
<p>Br^-1 + MnO4^-1 --> Br2 + Mn^2+</p>
<p>2Br^-1 --> Br2 + 2e-</p>
<p>MnO4^-1 + 8H+ + 5e- --> Mn^2+ 4H2O</p>
<p>Mn in MnO4-1 has an oxidation number of +7</p>
<p>10Br^-1 + 2MnO4^-1 + 16H+ --> 5Br2 + 2Mn^2+ + 8H2O</p>
<p>hmm, we were taught that in HS though..</p>
<ol>
<li>23.5 mL of a 0.0100 M Ca(OH)2 react with 15.0 ml of HCL, what is the molarity of HCL? HOw many grams of HCL theorhetically exist?</li>
</ol>
<p>Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O
Mols of Ca(OH)2 = 23.5 * 0.01
1 mol of Ca(OH)2 reacts with 2 mols of HCl
thus, 23.5 * 0.01 *2 = 15 * M
M = 0.03133 M</p>
<p>0.03133 * (1 + 35.5) = 1.143545 grams</p>
<p>should be correct unless i slipped up somewhere again LOL.</p>