Question

If a sample of 0.140 g of `KCN` is treated with an excess of `HCl`, how do you calculate the amount of `HCN` formed, in grams?

When potassium cyanide (`KCN`) reacts with acids, a deadly poisonous gas, hydrogen cyanide (`HCN`), is given off. Here is the equation: `KCN(aq) + HCl(aq) -> KCI(aq) + HCN(g)`

Answer 1

`"Moles of potassium cyanide"` `-=` `"Moles of hydrogen cyanide,"`

Explanation:

As with all problems of this type we write an equation to represent the stoichiometric equivalence:

`K^+""^(-)C-=N(aq) + HCl(aq) rarr H-C-=N(g) + K^+Cl^(-)(aq)`

And thus there is `1:1` equivalence.

`"Moles of potassium cyanide"` `-=` `(0.140*g)/(65.12*g*mol^-1)`

`=2.15xx10^-3*mol.`

And given the equivalence, we multiply this molar quantity by the molecular mass of hydrogen cyanide,

`=2.15xx10^-3*molxx27.03*g=58.1*mg.`

Hydrogen cyanide has been used as a rat poison, and as a human poison. It has the faint odour of almonds, not that you want to smell it.