PDA

View Full Version : reaction


haber101
Feb14-10, 09:01 PM
1.05 g MgO is added to 110.15ml HCL with an initial temperature of 21.5 degrees Celsius and a final temperature of 34.5 degrees Celsius. calculate the KJ of energy released for the reaction. assume that 4.18 J of energy is required to change the temperature of one mL of solution to one Celsius degree.


my thinking. q=mcdeltat

q=?
m=110.15ml
c=4.18kj
delta t=13

=(110.15)(4.18)(13)
=5985.55Kj
=6.0x10^3

Borek
Feb15-10, 02:31 AM
Your thinking is not incorrect, but you are incredibly cruel to your units. Mass is not in ml, and your use of k is completely off.

--
chemical calculators (http://www.chembuddy.com) - buffer calculator (http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator), concentration calculator (http://www.chembuddy.com/?left=CASC&right=concentration_and_solution_calculator)
www.titrations.info (http://www.titrations.info) - all about titration (http://www.titrations.info) methods

haber101
Feb15-10, 08:12 AM
Your thinking is not incorrect, but you are incredibly cruel to your units. Mass is not in ml, and your use of k is completely off.

--
chemical calculators (http://www.chembuddy.com) - buffer calculator (http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator), concentration calculator (http://www.chembuddy.com/?left=CASC&right=concentration_and_solution_calculator)
www.titrations.info (http://www.titrations.info) - all about titration (http://www.titrations.info) methods

thanks, so how about i transfer the 110.15ml to grams which is 110.15g and then do q=mc\deltat
q=(110.15g)(4.18J/g^oC)(13^oC)
q=5985.551 J = 598 KJ
q=5.9x10^3 J

so the answer is q= 598 KJ or q=5.985x10^3 J ??

better?

also how can I calculate KJ/mol of MgO from this? has to do with the molar mass in guessing? which for Mg is 24.31 and oxygen 16.0 so 40.31 so 5985/40.31=148.47 Kj/mol of MgO. did i do this correctly?

Borek
Feb15-10, 11:16 AM
q=(110.15g)(4.18J/g^oC)(13^oC)

Very nice and very correct.

q=5985.551 J = 598 KJ

Nice try - but wrong. kJ is 1000 Joules.

also how can I calculate KJ/mol of MgO from this? has to do with the molar mass in guessing? which for Mg is 24.31 and oxygen 16.0 so 40.31 so 5985/40.31=148.47 Kj/mol of MgO. did i do this correctly?

No. You need molar mass, but you have to use it to calculate number of moles of MgO. Again - if you would pay attention to units it would be obvious something is wrong.

--
chemical calculators (http://www.chembuddy.com) - buffer calculator (http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator), concentration calculator (http://www.chembuddy.com/?left=CASC&right=concentration_and_solution_calculator)
www.titrations.info (http://www.titrations.info) - all about titration (http://www.titrations.info) methods

haber101
Feb15-10, 11:38 AM
Very nice and very correct.



Nice try - but wrong. kJ is 1000 Joules.



No. You need molar mass, but you have to use it to calculate number of moles of MgO. Again - if you would pay attention to units it would be obvious something is wrong.

--
chemical calculators (http://www.chembuddy.com) - buffer calculator (http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator), concentration calculator (http://www.chembuddy.com/?left=CASC&right=concentration_and_solution_calculator)
www.titrations.info (http://www.titrations.info) - all about titration (http://www.titrations.info) methods

ok, thanks, i see so it must be 5.98KJ, makes sense

for the second part, hmmm how about if i did the amount of MgO which is given

(1.05g MgO)(1mol MgO)/(40.31 g MgO)=.0260

im not sure about this one

Borek
Feb15-10, 11:44 AM
As usual you omitted units in your answer, but I guess you mean 0.0260 moles of MgO?

You know 5.98 kJ (beware: small k, not capital K) was produced by reaction of 0.0260 moles of MgO. Now just combine these numbers so that the answer is in kJ/mol.

--
chemical calculators (http://www.chembuddy.com) - buffer calculator (http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator), concentration calculator (http://www.chembuddy.com/?left=CASC&right=concentration_and_solution_calculator)
www.titrations.info (http://www.titrations.info) - all about titration (http://www.titrations.info) methods

haber101
Feb15-10, 11:49 AM
As usual you omitted units in your answer, but I guess you mean 0.0260 moles of MgO?

You know 5.98 kJ (beware: small k, not capital K) was produced by reaction of 0.0260 moles of MgO. Now just combine these numbers so that the answer is in kJ/mol.

--
chemical calculators (http://www.chembuddy.com) - buffer calculator (http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator), concentration calculator (http://www.chembuddy.com/?left=CASC&right=concentration_and_solution_calculator)
www.titrations.info (http://www.titrations.info) - all about titration (http://www.titrations.info) methods

yes that is what i meant, my apologies

by combine do you just mean say that there is 5.98 kJ per 0.0260 moles of MgO

or 5.98kJ/0.0260 mol MgO= 230kJ/mol

Borek
Feb15-10, 12:05 PM
Looks OK :smile:

Note: I have not checked numbers, but they look reasonable.

--
ChemBuddy chemical calculators (http://www.chembuddy.com) - buffer calculator (http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator), stoichiometry calculator (http://www.chembuddy.com/?left=EBAS&right=equation-balancing-stoichiometry)
www.ph-meter.info (http://www.ph-meter.info) - ph meter (http://www.ph-meter.info), ph electrode (http://www.ph-meter.info/pH-electrode)

haber101
Feb15-10, 12:11 PM
Looks OK :smile:

Note: I have not checked numbers, but they look reasonable.

--
ChemBuddy chemical calculators (http://www.chembuddy.com) - buffer calculator (http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator), stoichiometry calculator (http://www.chembuddy.com/?left=EBAS&right=equation-balancing-stoichiometry)
www.ph-meter.info (http://www.ph-meter.info) - ph meter (http://www.ph-meter.info), ph electrode (http://www.ph-meter.info/pH-electrode)

i appreciate the help, thanks