Question

GIven, `2NH_3(g)+2CH_4(g)+3O_2(g)rarr2HCN(g)+6H_2O(g)`... ...what mass of hydrocyanic acid would result from a `203.6*g` mass of ammonia?

Answer 1

I gets..........over `300*g` of `HC-=N` given QUANTITATIVE reaction.

Explanation:

You have given the stoichiometric equation:

`2NH_3(g) + 2CH_4 + 3O_2(g) rarr 2HCN(g) + 6H_2O(g)`

`"Moles of ammonia"=(203.6*g)/(17.03*g*mol^-1)=11.9*mol`.

And now attend! The stoichiometry CLEARLY indicates that a given molar quantity of ammonia will give an EQUIMOLAR quantity of hydrogen cyanide. And given quantitative reaction, we should, with respect to hydrogen cyanide, get a mass of............

`11.9*molxx27.03*g*mol^-1=??g`

I think this is the old `"Andrussow Process"` which is performed to this day.