Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Lindsay on October 30, 2008, 05:46:45 PM
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Please I need your sincere help with this problems
Problem 1
How much heat is released when a mixture containing 10.0 g CS2 and 10.0 g Cl2
CS2 + 3Cl2 -------- S2Cl2 + CCl4
Delta H = -230 kJ
Answer
10.0 g CS2 X 1 mol CS2/76.15 g X 3 mol Cl2/1 mol CS2 = 0.39395929 Cl2
10.0 g Cl2 X 1 mol Cl2/ 70.90 g X 1 mol CS2/3 mol Cl2 = 0.04701457 CS2
LR
10.0 g Cl2 X 1 mol Cl2/ 70.90 g = 0.14104372 mol Cl2
Delta H = -230 kJ/3 mol Cl2 X 0.14104372 mol Cl2 = -10.8 kJ
Problem 2
NH4NO3----- N2 + O + 2H2O
Delta Heat= -30 Kcal
NH4NO3--2.5 Tons
How much water 0---100 C in tons
2.5 tons NH4NO3 X 2000 lb / 1 ton X 453.59 g/ 1lb = 1 mol/80.05 g
= 2.8 X 10 (4) mol NH4NO3
-30 Kcal/1mol NH4NO3 X 2.84 X 10 (4) mol NH4NO3
= 8.5 X 10 (5) Kcal X 1000 = 8.5 X 10 (8) cal
8.5 X 10 (8) cal/100 C X 1g C/1 cal X 1lb/453.59 g X 1Ton/2000 =9.25 Tons
Problem 3
2 Na + 2 H2O---- 2 NaOH + H2
Delta H = -3686 kJ
1.0 g 2Na
1.0 g 2H2O
How much heat???
1.0 g Na X 1 mol Na/22.99 g X 2 mol H2O/2 mol Na = 0.0434971727 H2O
1.0 g H2O X 1 mol H2O/ 18.02 g X 2 mol Na/2 mol H2O = 0.0554938957 Na
LR
1.0 g Na X 1 mol Na/ 22.99 g = 0.0434971727 mol Na
Delta H = -3686 kJ/2 mol Na X 0.0434971727 mol Na = -80 kJ
Problem 4
H2 + .5 O2 ----- H2O
Delta H = -58 Kcal
O2 629,340 Kg
H2 106,261 Kg
a. How many calories are emitted from launch?
b. If I were to take energy to heat up a pan...how much water can I heat from 0 to 100 C?
Answer
629,340 kg X 1000 = 6.2934 X 10 (8) g O2 * 1 mol O2/16.00 g O2 * 1 mol H2/ .5 mol O2 = 7.86675 X 10 (7) mol H2
106,261 kg X 1000 = 1.06261 X 10 (8) g H2 * 1 mol H2/2.016 g H2 * .5 mol O2/ 1 mol H2 = 2.6354414 X 10 (7)
Limited Reactant
106,261 kg X 1000 = 1.06261 X 10 (8) g H2 * 1 mol H2/2.016 g H2
= 5.2708829 X 10 (7) H2
-58 Kcal/1 mol H2 * 5.2708829 X 10 (7) H2 = -3.057112103 X 10(9) kcal X 1000 = -3.0571121 X 10 (12) cal
-3.06 X 10 (12) cal * 1g C/1 cal X 1/100 C X 1lb/453.59 g X 1 ton/ 2000 lb = -33699 Tons of Water
I'm not sure about the last problem
Are these answers correct?
Please advice
Thanks
Lindsay
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Here I'm posting it again. Somehow smily faces appeared on the first post
Please I need your sincere help with this problems
Problem 1
How much heat is released when a mixture containing 10.0 g CS2 and 10.0 g Cl2
CS2 + 3Cl2 -------- S2Cl2 + CCl4
Delta H = -230 kJ
Answer
10.0 g CS2 X 1 mol CS2/76.15 g X 3 mol Cl2/1 mol CS2 = 0.39395929 Cl2
10.0 g Cl2 X 1 mol Cl2/ 70.90 g X 1 mol CS2/3 mol Cl2 = 0.04701457 CS2
LR
10.0 g Cl2 X 1 mol Cl2/ 70.90 g = 0.14104372 mol Cl2
Delta H = -230 kJ/3 mol Cl2 X 0.14104372 mol Cl2 = -10.8 kJ
Problem 2
NH4NO3----- N2 + O + 2H2O
Delta Heat= -30 Kcal
NH4NO3--2.5 Tons
How much water 0---100 C in tons
2.5 tons NH4NO3 X 2000 lb / 1 ton X 453.59 g/ 1lb = 1 mol/80.05 g
= 2.8 X 10 to the 4 mol NH4NO3
-30 Kcal/1mol NH4NO3 X 2.84 X 10 to the 4mol NH4NO3
= 8.5 X 10 to the 5 Kcal X 1000 = 8.5 X 10 to the 8 cal
8.5 X 10 to the 8 cal/100 C X 1g C/1 cal X 1lb/453.59 g X 1Ton/2000 =9.25 Tons
Problem 3
2 Na + 2 H2O---- 2 NaOH + H2
Delta H = -3686 kJ
1.0 g 2Na
1.0 g 2H2O
How much heat???
1.0 g Na X 1 mol Na/22.99 g X 2 mol H2O/2 mol Na = 0.0434971727 H2O
1.0 g H2O X 1 mol H2O/ 18.02 g X 2 mol Na/2 mol H2O = 0.0554938957 Na
LR
1.0 g Na X 1 mol Na/ 22.99 g = 0.0434971727 mol Na
Delta H = -3686 kJ/2 mol Na X 0.0434971727 mol Na = -80 kJ
Problem 4
H2 + .5 O2 ----- H2O
Delta H = -58 Kcal
O2 629,340 Kg
H2 106,261 Kg
a. How many calories are emitted from launch?
b. If I were to take energy to heat up a pan...how much water can I heat from 0 to 100 C?
Answer
629,340 kg X 1000 = 6.2934 X 10 to the 8 g O2 * 1 mol O2/16.00 g O2 * 1 mol H2/ .5 mol O2 = 7.86675 X 10 to the 7 mol H2
106,261 kg X 1000 = 1.06261 X 10 to the 8 g H2 * 1 mol H2/2.016 g H2 * .5 mol O2/ 1 mol H2 = 2.6354414 X 10 to the 7
Limited Reactant
106,261 kg X 1000 = 1.06261 X 10 to the 8 g H2 * 1 mol H2/2.016 g H2
= 5.2708829 X 10 to the 7 H2
-58 Kcal/1 mol H2 * 5.2708829 X 10 to the 7 H2 = -3.057112103 X 10 to the 9 kcal X 1000 = -3.0571121 X 10 to the 12 cal
-3.06 X 10 to the 12 cal * 1g C/1 cal X 1/100 C X 1lb/453.59 g X 1 ton/ 2000 lb = -33699 Tons of Water
I'm not sure about the last problem
Are these answers correct?
Please advice
Thanks
Lindsay
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Problem 1 looks correct(step-wise)
Problem 2:
8.5 X 10 to the 8 cal/100 C X 1g C/1 cal X 1lb/453.59 g X 1Ton/2000 =9.25 Tons
Are you using m = Q/c :delta: T ? If yes then your stepwise is correct.
Problem 3
2 Na + 2 H2O---- 2 NaOH + H2
Na + H2O ----> NaOH + 1/2 H2 (Write it in per mole)
Check your working again.
Problem 4:
629,340 kg X 1000 = 6.2934 X 10 to the 8 g O2 * 1 mol O2/16.00 g O2 * 1 mol H2/ .5 mol O2 = 7.86675 X 10 to the 7 mol H2
106,261 kg X 1000 = 1.06261 X 10 to the 8 g H2 * 1 mol H2/2.016 g H2 * .5 mol O2/ 1 mol H2 = 2.6354414 X 10 to the 7
7 mol? It's different from your balanced equation. Your limiting reagent is H2?
you have done pretty good ;D, and usually if enthalpy is given, they are given in terms of per mole, so it would be good to write your stoichiometric ratio in the simplest term.
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Thanks Astrokel for your response.
Questions..
Why are you multiplying the equation on problem 3 by 2? I don’t understand..and don't you have to multiply the -3686 by 2 too?
Original 2 Na + 2 H2O---- 2 NaOH + H2
You changed it to Na + H2O ----> NaOH + 1/2 H2 (Write it in per mole)
Delta H = -3686 kJ
1.0 g 2Na
1.0 g 2H2O
How much heat???
1.0 g Na X 1 mol Na/22.99 g X 2 mol H2O/2 mol Na = 0.0434971727 H2O
1.0 g H2O X 1 mol H2O/ 18.02 g X 2 mol Na/2 mol H2O = 0.0554938957 Na
LR
1.0 g Na X 1 mol Na/ 22.99 g = 0.0434971727 mol Na
Original …. Delta H = -3686 kJ/2 mol Na X 0.0434971727 mol Na = -80 kJ
Changing it to ….. Delta H = -3686 kJ/1 mol Na X 0.0434971727 mol Na = -160 kJ
So the new result is -160 kJ
Problem 4
Is this problem wrong? I thought the Limited Reactant was H2.
Also, can I multiply the equation by 2 and get H2 + O2 ----- H2O?
Original problem
H2 + .5 O2 ----- H2O
Delta H = -58 Kcal
O2 629,340 Kg
H2 106,261 Kg
a. How many calories are emitted from launch?
b. If I were to take energy to heat up a pan...how much water can I heat from 0 to 100 C?
Answer
629,340 kg X 1000 = 6.2934 X 10 to the 8 g O2 * 1 mol O2/16.00 g O2 * 1 mol H2/ .5 mol O2 = 7.86675 X 10 to the 7 mol H2
106,261 kg X 1000 = 1.06261 X 10 to the 8 g H2 * 1 mol H2/2.016 g H2 * .5 mol O2/ 1 mol H2 = 2.6354414 X 10 to the 7 O2
Limited Reactant
106,261 kg X 1000 = 1.06261 X 10 to the 8 g H2 * 1 mol H2/2.016 g H2
= 5.2708829 X 10 to the 7 H2
-58 Kcal/1 mol H2 * 5.2708829 X 10 to the 7 H2 = -3.057112103 X 10 to the 9 kcal X 1000 = -3.0571121 X 10 to the 12 cal
-3.06 X 10 to the 12 cal * 1g C/1 cal X 1/100 C X 1lb/453.59 g X 1 ton/ 2000 lb = -33699 Tons of Water
Thanks,
Lindsay
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I went ahead and did problem 4 again. This time I got a different answer. Please help
H2 + .5 O2 ----- H2O
I multiply the equation by 2 and get
2H2 + O2-----2H2O
Delta H = -58 Kcal ( Do I have to multiply this by two too?)
O2 629,340 Kg
H2 106,261 Kg
a. How many calories are emitted from launch?
b. If I were to take energy to heat up a pan...how much water can I heat from 0 to 100 C?
Answer
629,340 kg X 1000 = 6.2934 X 10 to the 8 g O2 * 1 mol O2/32.00 g O2 * 2 mol H2O/ 1 mol O2 = 3.9333750 X 10 to the 7 mol H2O
106,261 kg X 1000 = 1.06261 X 10 to the 8 g H2 * 1 mol H2/2.016 g H2 * 2 mol H2O/ 2 mol H2 = 5.2708829 X 10 to the 7 H2O
Limited Reactant is O2
-58 Kcal/1 mol O2 * 3.9333750 X 10 to the 7 = -2.281357500 X 10 to the 9 kcal X 1000 = -2.2813575 X 10 to the 12 cal
-2.28 X 10 to the 12 cal * 1g C/1 cal X 1/100 C X 1lb/453.59 g X 1 ton/ 2000 lb = -25,000 Tons of Water
Is this answer correct?
Thanks,
Lindsay
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Hi Lindsay ;D,
Why are you multiplying the equation on problem 3 by 2? I don’t understand..and don't you have to multiply the -3686 by 2 too?
Original 2 Na + 2 H2O---- 2 NaOH + H2
You changed it to Na + H2O ----> NaOH + 1/2 H2 (Write it in per mole)
You are aware that enthalpy change is mostly expressed in kJ mol-1 and that when Delta H is given in your question, it has an 'hidden agenda' that it is in per mole. If the stoichiometric ratio is what you have written originally then it is not in per mol as there will be 2 moles of sodium.
If you look at the definitions, for example, standard enthalpy of formation is the enthalpy change when 1 mole of compound is formed from its constituent elements under standard conditions.
H2 + .5 O2 ----- H2O
I multiply the equation by 2 and get
2H2 + O2-----2H2O
Delta H = -58 Kcal ( Do I have to multiply this by two too?)
Well, the original equation given to you is already balanced and it means that per mole of hydrogen gives you this enthalpy change of -58 Kcal. Why cannot work with the original equation?
But your answer to a is correct oxygen is the limiting. As for B, what formula are you using?
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Thanks for your response.
1g degree celsius per 1 calorie. Is that right? Or should I use;
Q=s* m* Change in Temperature
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it looks fine, except that you can drop your negative sign in b. good work ;D