# 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 ($K C N$) reacts with acids, a deadly poisonous gas, hydrogen cyanide ($H C N$), is given off. Here is the equation: $K C N \left(a q\right) + H C l \left(a q\right) \to K C I \left(a q\right) + H C N \left(g\right)$

Jan 8, 2017

$\text{Moles of potassium cyanide}$ $\equiv$ $\text{Moles of hydrogen cyanide,}$

#### 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.

$\text{Moles of potassium cyanide}$ $\equiv$ $\frac{0.140 \cdot g}{65.12 \cdot g \cdot m o {l}^{-} 1}$

$= 2.15 \times {10}^{-} 3 \cdot m o l .$

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

$= 2.15 \times {10}^{-} 3 \cdot m o l \times 27.03 \cdot g = 58.1 \cdot m g .$

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.