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<div class="moz-cite-prefix">On 6/12/21 10:20 AM, <br>
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cite="mid:27c04022f52e41898ca1be38141fb3e2@SA1PR01MB6528.prod.exchangelabs.com">
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<div>Good morning Dr. Pounds, <span></span></div>
<div dir="ltr">I’m confused about how I use the density of
NaCl and water to find the mass of the solution. Do I add
the densities of the products together then multiply by
the mass of the reactants or is there a ratio I’m supposed
to know? And when we find the Enthalpies the equations are
balanced so does that mean the moles are 1?</div>
<div dir="ltr">Thank you for your help<span></span></div>
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<div class="moz-cite-prefix">So in the following experiments...</div>
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<div class="moz-cite-prefix">NaOH (aq) + HCl (aq) -> H2O (l) +
NaCl (aq)</div>
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<div class="moz-cite-prefix">NaOH (s) + HCl (aq) -> H2O (l) +
NaCl (aq)</div>
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<div class="moz-cite-prefix">You immediately form H2O and NaCl
because sodium hydroxide and hydrochloric acids are strong
electrolytes. To computer the mass of the solution for the first
reaction above you assume that both volumes (for HCl and NaOH) are
additive and use the density of the NaCl solution shown on the
report form.</div>
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<div class="moz-cite-prefix">For the second reaction you could make
the same assumption that the final solution will be the same NaCl
as for the first experiment above an assume that the volume of the
NaOH pellets is negligible. <br>
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<div class="moz-cite-prefix">So in both cases the mass of the water
would simply be the total volume of each solution multiplied by
the density given in the report form.</div>
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<div class="moz-cite-prefix">The heat in each case would be
<math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>q</mi><mo>=</mo><mi>m</mi><msub><mi>c</mi><mi>s</mi></msub><mi
mathvariant="normal">Δ</mi><mi>T</mi></mrow><annotation
encoding="TeX">q=m c_s \Delta T</annotation></semantics></math>
. <br>
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<div class="moz-cite-prefix">However the ENTHALPY in each case is
based of the heat and the moles of NaOH consumed in the reaction.</div>
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<div class="moz-cite-prefix">For the reaction...</div>
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<div class="moz-cite-prefix">NaOH (s) -> Na+ + OH-</div>
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<div class="moz-cite-prefix">Just compute the moles of NaOH by
dividing the mass of NaOH you used by the molar mass of NaOH.
Then divide the heat by that number of moles to get the enthalpy.</div>
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<div class="moz-cite-prefix">For the reaction</div>
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<div class="moz-cite-prefix">NaOH (s) + HCl (aq) -> H2O (l) +
NaCl (aq)</div>
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<div class="moz-cite-prefix">You do the exact same thing.</div>
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<div class="moz-cite-prefix">For the reaction <br>
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<div class="moz-cite-prefix">NaOH (aq) + HCl (aq) -> H2O (l) +
NaCl (aq)</div>
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<div class="moz-cite-prefix">You have to compute the moles of NaOH
you used by multiplying the volume of NaOH in liters by the
molarity of the NaOH. Then divide the heat by that number of
moles to get the enthalpy. <br>
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<div class="moz-cite-prefix">Remember -- enthalpy is an extensive
property so we are having to base our calculation off the amount
we used. If we were to use and entire mole of NaOH, then you
would use "1" in your calculations.<br>
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<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Director of the Computational Science Program
Mercer University, Macon, GA 31207 (478) 301-5627
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