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?

Question

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

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Answer 1 · 2017-01-08T03:26:25

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

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.