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Part A, the reaction of Mg(s) with excess HCI Mass of Mg (g) Moles Mg (mole) g solution Ccalorimeter (J/°C (g solution x 4.184) ?T (from graph) q<sub>p</sub> (J) = C<sub>calorimeter</sub>?T Use sig. figs. rule: express answer in scientific notation! q(kJ) ?H<sub>reaction</sub> (kJ/mol) = q<sub>p</sub>/(moles Mg) Average ?H (kJ/mol) Trial 1 0.41 200 g 200 x 4.184 = 836.8 Trial 2 Trial 3 (if time available) 0.405 200 g 200 x 4.184 = 836.8

          Part A, the reaction of Mg(s) with excess HCI
Mass of Mg (g)
Moles Mg (mole)
g solution
Ccalorimeter (J/°C
(g solution x 4.184)
?T (from graph)
q<sub>p</sub> (J) = C<sub>calorimeter</sub>?T
Use sig. figs. rule: express
answer in scientific notation!
q(kJ)
?H<sub>reaction</sub> (kJ/mol) =
q<sub>p</sub>/(moles Mg)
Average ?H (kJ/mol)
Trial 1
0.41
200 g
200 x 4.184 = 836.8
Trial 2
Trial 3 (if time
available)
0.405
200 g
200 x 4.184 = 836.8
        
Show more…
Part A, the reaction of Mg(s) with excess HCI
Mass of Mg (g)
Moles Mg (mole)
g solution
Ccalorimeter (J/°C
(g solution x 4.184)
?T (from graph)
q<sub>p</sub> (J) = C<sub>calorimeter</sub>?T
Use sig. figs. rule: express
answer in scientific notation!
q(kJ)
?H<sub>reaction</sub> (kJ/mol) =
q<sub>p</sub>/(moles Mg)
Average ?H (kJ/mol)
Trial 1
0.41
200 g
200 x 4.184 = 836.8
Trial 2
Trial 3 (if time
available)
0.405
200 g
200 x 4.184 = 836.8

Added by Jennifer K.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Part Athe reaction of Mg(swith excess HCl Trial1 Trial2 Trial3if time available) Mass of Mgg 1h-0 0405 Moles Mgmole g solution 2009 2009 CcalorimterJ/C g solution 4.184 200X4184836.8 T(from graph qpJ=CcalorimeterT Use sig.figs.rule:express answer in scientific notation! 200x4:184=836.8 q(k) HreactionkJ/mol= q/moles Mg Average AH (kJ/mol)
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Transcript

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00:01 Okay, in this problem we're asked to consider this reaction and we're carrying out, carrying it out as described in experiment and we have 0 .636 grams of magnesium.
00:14 We determined we find our heat for the reaction to equal 5 .12 kilojoules and we're asked to, no, that's not find, we found, okay, and now we have to calculate kilojoules per mole of magnesium.
00:52 Okay, easy peasy.
00:54 0 .636 grams of magnesium.
01:01 I'm actually gonna do it like this.
01:03 5 .12 kilojoules per 0 .636 grams of magnesium.
01:18 So now let me get to my periodic table so i get the right molar mass, 24 .31 and this will equal, let me do this, 5 .12 times 24 .31 divide 0 .636 and i get 195 .7 kilojoules per mole and let me go look at my sig figs for my periodic table.
01:55 I have three and three.
01:57 So i'm gonna report this as 196 kilojoules per mole of magnesium.
02:12 Okay, and this will be negative, i think...
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