0:02 so here's an example of calculating our
0:05 Q for a reaction um and calculating
0:08 enthalpies uh for and how to do it this
0:10 so what we're doing is we are going to
0:13 react zinc with some hydrochloric acid
0:16 and we know that we have
0:20 3.14 K of heat are produced so as a
0:23 reminder when we say that the heat is
0:25 produced this is
0:28 exothermic of a reaction and then we
0:31 want to calculate the change of enthalpy
0:34 per mole of zinc for the reaction so we
0:38 want to calculate the Delta H for this
0:41 reaction and this Delta H is based on
0:44 the reaction itself it could be per mole
0:48 of zinc it could be per two moles of HCL
0:50 so we use our stochiometry to think
0:55 about this Delta H so first of all our Q
0:59 of the reaction is going to equal - 3.14
1:02 k where did I get that because this is
1:04 exothermic and whenever we have an
1:07 exothermic reaction remember that we
1:13 have a goes to B plus heat so heat for
1:17 the reaction is leaving the reaction so
1:22 that is a negative Q so this is our Q
1:25 reaction um and then we have remember
1:27 what we're also looking at is we have
1:30 two reactants here so when we have two
1:32 reactants we have to think limiting
1:34 reagents so which one of those is our
1:37 limiting reagents so first of all we
1:40 have 60.0 milliliters we're going to
1:46 liters and then we're going to find how
1:57 have so and then when we do this we need
2:01 to convert from moles of hydroch acid to
2:05 zinc so there's one zinc for every two
2:08 hcls and if we calculate that out we end
2:10 up getting
2:14 0.0225 moles of
2:17 zinc so that works out but then we have
2:18 to see how many moles of zinc do we
2:22 actually have so we can start with 1.34
2:23 grams of
2:26 zinc we can go ahead and convert that
2:28 using our molecular weight so it's one
2:34 mole for every 65.3 n g of zinc and we
2:40 0.0204 and I'll carry an extra digit
2:44 nine of of uh moles of
2:47 zinc so the hydrochloric acid requires
2:50 more zinc than we have and then the zinc
2:52 is there's actually not enough so if we
2:54 think about
2:56 025 uh
2:58 0225 and
3:02 0204 we know that we have less zinc than
3:06 we need so that will be our limiting
3:09 reagent so we can find out that we have
3:12 this many uh moles of zinc then we go
3:15 back to our problem we're trying to have
3:19 our Delta H and that's going to be our Q
3:21 of the reaction which we've already got up
3:22 up
3:27 here and then how many moles of zinc do
3:30 we have and then that value is right
3:33 here so we're going to have a minus 3.4 K
3:35 K
3:39 314 divided
3:42 0.0204 and I'll carry in an extra Sig
3:46 fig moles of zinc and this ends up be minus
3:48 minus 153
3:50 153
3:52 kles uh per
3:57 mole of zinc and that is our final
