the specific heat of metal is 0.11 and its equivalent weight is 18.61 therefore its exact atomic weight is
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Home»Forum»Discuss with colleagues and IITians»chemistry-easy-qstn-6
i need help with these few questions ..thnks :D
1.During preparation of ammonia by Habers process , 50 L of H2 and 30 L of N2 are mixed . The yield of NH3 is 60 % . Find the composition of gaseuos mixtures.
2. The specific heat of metal is 0.11cal/g and its equivalent weight is 18.61 .what will be its exact atomic weight ?
3.which solution will have the lowest molar concentraion ?
a. 1.0 N HCl
b. 0.4N H2SO4
c.0.1N Na2CO3
5 years ago
Answers : (1)
1 N2 + 3H2 = 2 NH3 balanced equation.
One mole of N2 reacts with 3 moles of H2
The ratio between these two gases by volume is also the same as the ratio between reacting moles.
so one liter of H2 reacts completely with 3 liters of H2
therefore 30 liters of N2 requires 90 liters of H2 for complete reaction
so you are short on H2. So H2 is the limiting reactant and will determine how much NH3 you can produce.
Use the balanced equation as a math platform. Over the 3H2 enter 50 Liters and below the same enter its coefficient of 3 moles.
Over the NH3 enter X liters and below the same enter its coefficient of 2 Moles.
Now cross multiply through the equation and solve for X Liters of NH3. I got 33.3 Liters. Of course this would be the maximum yield possible in this limiting reactant situation. But if you only get 60% of this, you get only .60 times 33.3 liters = 19.98 Liters of NH3
now use the balanced equation again as a math platform to find the number of liters of H2 and N2 that were actually involved in the reaction.
Over the NH3 enter l9.98 liters and below the same enter its coefficient of 2 moles
and over the N2 and also the H2 enter X liters and below each enter their coefficients of one and three.
Cross multiply again for both X liters N2 and X liters H2. I got X = 9.99 Liters N2 used
and X = 30 liters H2 used
H2 remaining 50 Liters minus 30 Liters = 20 Liters H2 remaining
N2 remaining 30Liters minus 9.99 Liters = about 20 Liters N2 remaining
and of course close to 20 liters of NH3 was produced. I hope I interpreted the 60% yield correctly.
2. Here is a way to find the approximate atomic wt.
Sp Heat times atomic mass = 3R
.11 calories times 4.184 joules per calorie = .46 joules per gram specific heat.
When your specific heat is in joules per gram, the value of R = 8.31
so .46 joules/gm times atomic mass = 3 times 8.31 or 24.93
.46 (X) = 24.93
X = 54.2 approx atomic wt.
Approx atomic wt over equivalent wt = oxidation number
54.2 over l8.61 = 2.91, rounds off to three
Oxidation number times equivalent wt. = exact atomic wt.
3 times l8.61 = 55.8 turns our to be iron, Fe
3. .10 Normal HCl divided by oxidation # of H which is one, = .10 Molar
.40 Normal H2SO4 divided by total oxidation # of H2 which is 2 = .20 Molar
.10 Normal Na2CO3 divided by total positive oxidation number of Na2 which is 2 = .05 Molar
so Na2CO3 has the lowest molar concentration
Normality over total positive oxid. # = Molarity
Molarity times total positive oxid # = Normality
approx atomic mass =
0
Home»Forum»Discuss with colleagues and IITians»chemistry-easy-qstn-6
i need help with these few questions ..thnks :D
1.During preparation of ammonia by Habers process , 50 L of H2 and 30 L of N2 are mixed . The yield of NH3 is 60 % . Find the composition of gaseuos mixtures.
2. The specific heat of metal is 0.11cal/g and its equivalent weight is 18.61 .what will be its exact atomic weight ?
3.which solution will have the lowest molar concentraion ?
a. 1.0 N HCl
b. 0.4N H2SO4
c.0.1N Na2CO3
5 years ago
Answers : (1)
1 N2 + 3H2 = 2 NH3 balanced equation.
One mole of N2 reacts with 3 moles of H2
The ratio between these two gases by volume is also the same as the ratio between reacting moles.
so one liter of H2 reacts completely with 3 liters of H2
therefore 30 liters of N2 requires 90 liters of H2 for complete reaction
so you are short on H2. So H2 is the limiting reactant and will determine how much NH3 you can produce.
Use the balanced equation as a math platform. Over the 3H2 enter 50 Liters and below the same enter its coefficient of 3 moles.
Over the NH3 enter X liters and below the same enter its coefficient of 2 Moles.
Now cross multiply through the equation and solve for X Liters of NH3. I got 33.3 Liters. Of course this would be the maximum yield possible in this limiting reactant situation. But if you only get 60% of this, you get only .60 times 33.3 liters = 19.98 Liters of NH3
now use the balanced equation again as a math platform to find the number of liters of H2 and N2 that were actually involved in the reaction.
Over the NH3 enter l9.98 liters and below the same enter its coefficient of 2 moles
and over the N2 and also the H2 enter X liters and below each enter their coefficients of one and three.
Cross multiply again for both X liters N2 and X liters H2. I got X = 9.99 Liters N2 used
and X = 30 liters H2 used
H2 remaining 50 Liters minus 30 Liters = 20 Liters H2 remaining
N2 remaining 30Liters minus 9.99 Liters = about 20 Liters N2 remaining
and of course close to 20 liters of NH3 was produced. I hope I interpreted the 60% yield correctly.
2. Here is a way to find the approximate atomic wt.
Sp Heat times atomic mass = 3R
.11 calories times 4.184 joules per calorie = .46 joules per gram specific heat.
When your specific heat is in joules per gram, the value of R = 8.31
so .46 joules/gm times atomic mass = 3 times 8.31 or 24.93
.46 (X) = 24.93
X = 54.2 approx atomic wt.
Approx atomic wt over equivalent wt = oxidation number
54.2 over l8.61 = 2.91, rounds off to three
Oxidation number times equivalent wt. = exact atomic wt.
3 times l8.61 = 55.8 turns our to be iron, Fe
3. .10 Normal HCl divided by oxidation # of H which is one, = .10 Molar
.40 Normal H2SO4 divided by total oxidation # of H2 which is 2 = .20 Molar
.10 Normal Na2CO3 divided by total positive oxidation number of Na2 which is 2 = .05 Molar
so Na2CO3 has the lowest molar concentration
Normality over total positive oxid. # = Molarity
Molarity times total positive oxid # = Normality
approx atomic mass =
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