Chem mole calculation...plz help

[xim7007]

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?

[Deadly_king]

 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

[$H00t!N& $t@r]

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?

[Deadly_king]

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

[Arthur Bon Zavi]

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

[Deadly_king]

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

[Arthur Bon Zavi]

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!

[Deadly_king]

Yes, yes! I misunderstood the Q...dont mind that!

Its alright bro......it happens!

Just be careful next time