Question

When potassium cyanide (KCN) reacts with acids, a deadly poisonous gas, hydrogen cyanide (HCN), is given off: KCN(aq) + HCl(aq) → HCN(g) + KCl(aq) If a sample of 0.420 g of KCN is treated with an excess of HCl, calculate the amount of HCN formed in grams.

Answer 1

Answer: The mass of hydrogen cyanide formed is 0.17 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

Given mass of KCN = 0.420 g

Molar mass of KCN = 65.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of KCN}=\frac{0.420g}{65.12g/mol}=0.0064mol$

The given chemical equation follows:

$KCN(aq.)+HCl(aq.)\rightarrow HCN(g)+KCl(aq.)$

As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.

Thus, potassium cyanide is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of potassium cyanide produces 1 mole of hydrogen cyanide.

So, 0.0064 moles of potassium cyanide will produce = $\frac{1}{1}\times 0.00064=0.0064mol$ of hydrogen cyanide

Now, calculating the mass of hydrogen cyanide from equation 1, we get:

Molar mass of HCN = 27.02 g/mol

Moles of HCN = 0.0064 moles

Putting values in equation 1, we get:

$0.0064mol=\frac{\text{Mass of HCN}}{27.02g/mol}\\\\\text{Mass of HCN}=0.17g$

Hence, the mass of hydrogen cyanide formed is 0.17 grams