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)#

1 Answer
Jan 8, 2017

Answer:

#"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.