Calculate the amount of ni needed in the mond's process if co used in this process is obtained through a process in which 6g of carbon is mixed with 44
g Co2
Answers
Answer:
14.75 grams of Nickel will be needed in Mond's process.
Explanation:
6 grams of carbon reacts with 44 grams of carbon dioxide which is being used in the reaction :
Ni + 4 CO - - - - - - > Ni ( CO )4
As,
C + CO2 - - - - - > 2 CO ---- 28 grams
1/2 mole : 6g 44 g
22 grams of carbon dioxide will be required.
Now, in another reaction:
Ni + 4 CO - - - - - - > Ni ( CO )4
1/4 mole: 28 grams
1/4 mole of Ni will be required that can be calculated :
Molecular mass of Ni = 59 = 59/4 = 14.75 grams of Nickel
Answer:
14.75
Explanation:
C+CO
2
→2CO
1 mole of C reacts with 1 mole of CO
2
, .i.e., 12 g of C reacts with 44 g of CO
2
but C is present in less amount and hence, C is the limiting reagent.
12 g C forms 2×28 g CO but 6 g form
12
(6×2×28)
, i.e., 28 g of CO.
Number of moles of CO =
28
28
=1
4×28 g of CO combine with 58.7 g of Ni.
So, 28 g of CO will combine with
4
58.7
=14.675 g of Ni.