1.) the chemical process of rusting is described by the following unbalanced equation:
Fe (s) + O2 (g) ---> Fe2O3 (s)
The maximum amount of rust that can be produced from 56 g F3 and 32g O2 is
a.) 160g or b) 80.0 show work, please
2.) For which one of the following acid solutions will 100.00mL of the acid solution exactly neutralize (react with) 50.0 mL of a .20 M Ba(OH)2 solution?
a.) .050 M HCI or b.) .10 M H2SO3 please show work
3.)A 20.0 mL sample of .200 M K2CO3 solution is added to 30.0 mL of .400 M Ba(NO3)2 solution. Barium carbonate precipitates. The concentration of barium ions, Ba2+, in solution after precipitation goes to completion is :
a) .150 M or b.) .160 M (show work)
4.)A 275 mL sample of O2 is collected over water at 60 degree Celsius . The total pressure of gas collected is 755 torr. If the water vapor is removed, what volume of O2 gas would be collected at STP? (The vapor pressure of water at 60 degree celsius is 149 torr.)
a.) 180 mL or b.) 224mL *please show work*
5.) Into a 3.00 liter container at 25 degree celsius are placed 1.23 moles of O2 gas and 3.20 moles of solid C(graphite) . If the carbon and oxygen react completely to form CO gas, what will be the final pressure in the container at 25 degree celsius?
a.) 5.01 atm or b.) 20.0 atm *show work please*
chemistry problems.help2
2009-09-30 4:49 pm
回答 (2)
2009-09-30 10:48 pm
✔ 最佳答案
1.)The answer is b).
4Fe(s) + 3O2(g) → 2Fe2O3(s)
No. of moles of Fe = 56/56 = 1 mol
No. of moles of O2 = 32/(16x2) = 1 mol
When 1 mol of Fe react, O2 needed = 1 x (3/4) = 0.75 mol
Hence, O2 is in excess, and Fe is the limiting reactant (completely reacts).
No. of moles of Fe reacted = 1 mol
Maximum no. of moles of Fe2O3 formed = 1 x (1/2) = 0.5 mol
Molar mass of Fe2O3 = 56x2 + 16x3 = 160 g/mol
Maximum mass of Fe2O3 formed = 0.5 x (56x2 + 16x3) = 80 g
2.
The answer is b).
a.) incorrect
Ba(OH)2 + 2HCl → BaCl2 + H2O
No. of moles of Ba(OH)2 added = 0.2 x (50/1000) = 0.01 mol
No. of moles of HCl added = 0.05 x (100/1000) = 0.005
When all Ba(OH)2 reacts, HCl needed = 0.01 x 2 = 0.02 mol
Hence, Ba(OH)2is in excess.
b.) correct
Ba(OH)2 + H2SO3 → BaSO3 + 2H2O
No. of moles of Ba(OH)2 used = 0.2 x (50/1000) = 0.01 mol
No. of moles of H2SO3 used = 0.1 x (100/1000) = 0.01 mol
When all Ba(OH)2 reacts, H2SO3 needed = 0.01 mol
Hence, both Ba(OH)2 and H2SO3 completely react.
3.)
The answer is b).
CO32-(aq) + Ba2+(aq) → BaCO3(s)
No. of moles of CO32- = 0.200 x (20.0/1000) = 0.004 mol
No. of moles of Ba2+ = 0.400 x (30.0/1000) = 0.012 mol
No. of moles of Ba2+ unreacted = 0.012 - 0.004 = 0.008 mol
Volume of the solution = (20 + 30)/1000 = 0.05 L
Concentration of Ba2+ in the solution = 0.008/0.05 = 0.160 M
4.)
The answer is a)
V1 = 275 mL, T1 = 60 + 273 = 333 K, P1 = 755 - 149 = 606 torr
V2 = ? mL, T2= 273 K, P2 = 760 torr
P1V1/T1 = P2V2/T2
(606)(275)/333 = 760V2/273
V2 = 180 mL
5.)
The answer is b.)
2C + O2 → CO
When O2 completely reacts, C needed = 1.23 x 2 = 2.46 mol
Hence, C is in excess, and O2 is the limiting reactant (completely reacts).
No. of moles of O2 reacted = 1.23 mol
No. of moles of CO2 formed = 1.23 x 2 = 2.46 mol
After reaction:
V = 3 L
n = 1.23 mol
R = 0.0821 atm L mol-1 K-1
T = 273 + 25 = 298 K
P = ? atm
PV = nRT
P = nRT/V = (2.46)(0.0821)(298)/3 = 20.0 atm
2009-09-30 6:40 pm
第1題)
4Fe + 3O2 → 2Fe2O3
Fe = 56 / 56 = 1mol
O2 = 32 / 32 = 1mol
先找limiting,
Fe = 1 / 4 = 0.25mol ;
O2 = 1 / 3 = 0.33mol
明顯Fe的mole較O2少,因此Fe是limiting.
表示要用Fe的mole推斷Fe2O3的mole.
Fe : Fe2O3 = 4 : 2
Fe2O3 = 1 / 2 = 0.5mol
0.5 * [ 2(56) + 3(16) ]
=80g
Ans = B
ps:好快到第2題,順便說說,如果我幫到你,希望你能盡快選我做最佳,因為yahoo知識太多人只問不選,幫了他又不選最佳,甚至把問題移除,真的很討厭+浪費我們的時間,如果這情形惡化,這裡不會再有人幫大家.
2009-09-30 10:48:16 補充：
第2題)
2HCl + Ba(OH)2 → BaCl2 + 2H2O
H2SO3 + Ba(OH)2 → BaSO3 + 2H2O
Ba(OH)2 = 0.2 * (50/1000)
= 0.01mol
所以,要剛好neutralization,acid的mole也要 = 0.01mol
字數太多,好快繼續
2009-09-30 10:48:35 補充：
HCl = 0.05 * (100/1000)
= 0.005mol
HCl:Ba(OH)2 = 2:1
HCl = 0.005 / 2 = 0.0025mol (mole明顯不相同)
H2SO3 = 0.1 * (100/1000)
= 0.01mol
H2SO3:Ba(OH)2 = 1:1
H2SO3 = 0.01mol
因此答案 = B
2009-09-30 10:58:53 補充：
第3題)
K2CO3(aq) + Ba(NO3)2(aq) → BaCO3(s) + 2KNO3(aq)
K2CO3 = 0.2 * (20/1000)
= 0.004mol
Ba(NO3)2 = 0.4 * (30/1000)
= 0.012mol
太多字
2009-09-30 10:59:04 補充：
K2CO3 = limiting,
即Ba(NO3)2會用剩,剩餘的Ba(NO3)2:
0.012 - 0.004 = 0.008mol
reaction後,solution的volume = 20+30 = 50ml
Ba(NO3)2 = 0.008 / (50/1000)
= 0.16M
所以答案 = B
2009-09-30 14:13:05 補充：
第4&5題我不肯定,好像未學過(?)
為免誤人子弟,都係交比其他高手補答^^"
2009-09-30 18:44:09 補充：
都估到係老爺子解答^^
因為我既化學程度只去到會考(升不到中六,沒有AL程度),
所以搞到虎頭蛇尾,sorry xdd
ps:感謝老爺子,你是我的偶像=)
4Fe + 3O2 → 2Fe2O3
Fe = 56 / 56 = 1mol
O2 = 32 / 32 = 1mol
先找limiting,
Fe = 1 / 4 = 0.25mol ;
O2 = 1 / 3 = 0.33mol
明顯Fe的mole較O2少,因此Fe是limiting.
表示要用Fe的mole推斷Fe2O3的mole.
Fe : Fe2O3 = 4 : 2
Fe2O3 = 1 / 2 = 0.5mol
0.5 * [ 2(56) + 3(16) ]
=80g
Ans = B
ps:好快到第2題,順便說說,如果我幫到你,希望你能盡快選我做最佳,因為yahoo知識太多人只問不選,幫了他又不選最佳,甚至把問題移除,真的很討厭+浪費我們的時間,如果這情形惡化,這裡不會再有人幫大家.
2009-09-30 10:48:16 補充：
第2題)
2HCl + Ba(OH)2 → BaCl2 + 2H2O
H2SO3 + Ba(OH)2 → BaSO3 + 2H2O
Ba(OH)2 = 0.2 * (50/1000)
= 0.01mol
所以,要剛好neutralization,acid的mole也要 = 0.01mol
字數太多,好快繼續
2009-09-30 10:48:35 補充：
HCl = 0.05 * (100/1000)
= 0.005mol
HCl:Ba(OH)2 = 2:1
HCl = 0.005 / 2 = 0.0025mol (mole明顯不相同)
H2SO3 = 0.1 * (100/1000)
= 0.01mol
H2SO3:Ba(OH)2 = 1:1
H2SO3 = 0.01mol
因此答案 = B
2009-09-30 10:58:53 補充：
第3題)
K2CO3(aq) + Ba(NO3)2(aq) → BaCO3(s) + 2KNO3(aq)
K2CO3 = 0.2 * (20/1000)
= 0.004mol
Ba(NO3)2 = 0.4 * (30/1000)
= 0.012mol
太多字
2009-09-30 10:59:04 補充：
K2CO3 = limiting,
即Ba(NO3)2會用剩,剩餘的Ba(NO3)2:
0.012 - 0.004 = 0.008mol
reaction後,solution的volume = 20+30 = 50ml
Ba(NO3)2 = 0.008 / (50/1000)
= 0.16M
所以答案 = B
2009-09-30 14:13:05 補充：
第4&5題我不肯定,好像未學過(?)
為免誤人子弟,都係交比其他高手補答^^"
2009-09-30 18:44:09 補充：
都估到係老爺子解答^^
因為我既化學程度只去到會考(升不到中六,沒有AL程度),
所以搞到虎頭蛇尾,sorry xdd
ps:感謝老爺子,你是我的偶像=)
參考: 第2題待續, 第3題待續, 第4題待續
收錄日期: 2021-04-30 12:59:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090930000051KK00236