The amount of CO needed per kg of Ni in Mond's process given below is [At.wt.Ni=58]\(Ni+4CO \rightarrow Ni(CO)_{4}\)
1) 1.93kg
2) 1.52kg
3) 1.22kg,
4) 1.0kg

Calculate the amount of Ni needed in the Mond's process given below Ni+4CO rarr Ni(CO)_(4) If CO used in this process is obtained through a process, in which 6 g of carbon is mixed with 4 g CO_(2).

1 kg wt is equal to

1 kg wt is equal to

Formation of volatile Ni(CO)_(4) and then its subsquent decomposition into Ni and CO makes basis of Mond's process: Ni+4CO overset(T_(1))to Ni(CO)_(4) overset(T_(2))to Ni 4CO_(3) T_(1) and T_(2) are:

The geometries of Ni(CO)_(4) and Ni(PPh_(3))_(2)CI_(2) are .

How many kg of CaCo_(3) [Mol. Wt = 100g mol e^(-1)] is needed to produce 336 kg of CaO [Mol wt = 56g mol^(-1)] if % yield of the reaction given is 60%. CaCO_(3)(s) rightarrow CaO(s) + CO_(2)(g)

[Ni(CN)_(4)]^(2-) and [Ni(CO)_(4)] have