IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: xim7007 on August 25, 2010, 07:19:21 pm
-
When nitrogen reacts with hydrogen in the Haber Process only 17% of the nitrogen is
converted to ammonia. What volume of nitrogen and what volume of hydrogen would be
needed to produce 1 tonne of ammonia?
-
N2 + 3H2 ----> 2NH3
According to the equation 2 moles of nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia.
2 moles of ammonia is equivalent to 34g.( Mass = No of moles x Mr )
To produce two moles of ammonia 3(24)dm3 of hydrogen and 24dm3 of nitrogen is required.
first convert 1 tonne into 106grams
therefore in order to produce 1 tonne of ammonia.........3(24)/34*106 of hydrogen is needed.
Ans : 2.16 x 106dm3 of hydrogen.
same calculation applies for nitrogen from which you will obtain 24/34*106. This is equal to 7.06 x 105dm3.
Since this represents only 17%.........it implies that the total amount of nitrogen required will be (7.06 x 105)/17 x 100 which equals 4.15 x 106dm3.
Volume of hydrogen required : 2.16 x 106dm3
Volume of nitrogen required : 4.15 x 106dm3
Hope i've been of help :)
-
helllo ppl
i have some questions on the mole concept.. can someone please help.
1)Consider the combustion of pentanol: 2C5H11OH + 15O2 ---> 10CO2 + 12H2O
(a) How many grams of carbon dioxide are formed for each mole oxygen consumed?
(b) How many grams of carbon dioxide are formed for each gram of pentanol burned?
2) What mass of iron (iii) oxide will be produced by the complete oxidation of iron?
-
helllo ppl
i have some questions on the mole concept.. can someone please help.
1)Consider the combustion of pentanol: 2C5H11OH + 15O2 ---> 10CO2 + 12H2O
(a) How many grams of carbon dioxide are formed for each mole oxygen consumed?
(b) How many grams of carbon dioxide are formed for each gram of pentanol burned?
2) What mass of iron (iii) oxide will be produced by the complete oxidation of iron?
2C5H11OH + 15O2 -----> 10CO2 + 12H2O
a) According to the equation :
15 moles of oxygen are required to produce 10 mole of carbon dioxide
15 moles of oxygen are required to produce (10 x44)g of carbon dioxide ( Mr of CO2 = 44)
Therefore 1 mole of oxygen will produce (10 x 44)/15 = 29.3g of CO2
b) According to the equation :
2 moles of pentanol burns to form 10 moles of carbon dioxide
(2 x 88)g of pentanol burns to form (10 x 44)g of carbon dioxide (Mr of pentanol = 83)
1g of pentanol will therefore burn to form (10 x 44) / (2 x 88) = 2.5g of CO2
2) 4Fe + 3O2 ----> 2Fe2O3
According to the equation :
For complete oxidation to take place, 4 moles of Fe reacts with 3 moles of O2 to form 2 moles of Fe2O3
1 mole of Fe reacts with 3/4 moles of O2 to form 1/2 moles of Fe2O3
1 mole of Fe reacts with 3/4 moles of O2 to form (1/2 x 159.6)g of Fe2O3
Mr of Fe2O3 = 159.6
Mass of iron (iii) oxide formed upon complete combustion of 1 mole of Fe = (159.6/2) = 79.8g
-
Since this represents only 17%.........it implies that the total amount of nitrogen required will be (7.06 x 105)/17 x 100 which equals 4.15 x 106dm3.
i think this would be rather like this because u have exchanged the numbers:
7.06 x 105
----------- X 17 = 1.2002 x 1005
100
-
i think this would be rather like this because u have exchanged the numbers:
7.06 x 105
----------- X 17 = 1.2002 x 1005
100
Nope dude.
I did it right. let me get it clear for you.
17 % -------> 7.06 x 105 moles
100% -------> (7.06 x 105)/17 * 100 = 4.15 x 106moles.
Hope it helps :)
-
Nope dude.
I did it right. let me get it clear for you.
17 % -------> 7.06 x 105 moles
100% -------> (7.06 x 105)/17 * 100 = 4.15 x 106moles.
Hope it helps :)
Yes, yes! I misunderstood the Q...dont mind that!
-
Yes, yes! I misunderstood the Q...dont mind that!
Its alright bro......it happens! :)
Just be careful next time :P