Question

when potassium cyanide(kcn reacts with acids dealdy poisonous gas, hydrogen cyanide hcn is given of here is the equation. kcn(aq)+HCL(aq)> kcl(aq)+hcn (g) if a sample of 0.140g of kcn is treated with an excess of hcl calculate the amount of HCN formed in grams.

Answer 1

Question#19742

when potassium cyanide(kcn reacts with acids dealdy poisonous gas, hydrogen cyanide hcn is given of here is the equation. kcn(aq)+HCL(aq)> kcl(aq)+hcn (g) if a sample of 0.140g of kcn is treated with an excess of hcl calculate the amount of HCN formed in grams.

Solution:

According to the equation:

$$KCN + HCl = KCl + HCN$$

1 mol KCN produce 1 mol HCN

the molar mass of KCN is: $M(KCN) = (39.1 + 12 + 14) \times 1 \text{ g/mol} = 65.1 \text{ g/mol}$

the molar mass of HCN is: $M(HCN) = (1 + 12 + 14) \times 1 \text{ g/mol} = 27 \text{ g/mol}$

$$\left\{ \begin{array}{l} 65.1 \text{ g(KCN)} \Rightarrow 27 \text{ g(HCN)} \\ 0.14 \text{ g(KCN)} \Rightarrow x \text{ g(HCN)} \end{array} \right.$$ $$x = \frac{0.14 \times 27}{65.1} = 0.058 \text{ g}$$

Answer: 0.058 g HCN.