Question

Hydrogen cyanide gas can be produced by the following reaction:

2 CH₄(g) + 2 NH₃(g) + 3 O₂(g) $\rightarrow$ 2 HCN(g) + 6 H₂O(g)

A) what volume of oxygen gas (in L) is required to react completely to produce 550 L of hydrogen cyanide? Assume reactant and product volumes are measured at 300 kPa and 1200 degree Celcius.

B) what volume of water vapour (in L) will be produced during the production 255 L of hydrogen cyanide? Assume all reactant and product volumes are measured at 300 kPa and 1200 degree Celcius.

Answer 1

The reaction for production of Hydrogen cyanide gas is as shown below:

2CH49) + 2NH3(g) + 302(g) + 2HCN (9) + 6H2O(9)

A) Under given conditions of temperature pressure, the ratio of number of moles of two gases (as per reaction stoichiometry) is equal to the ratio of the volumes of two gases.

3 mol O2 = 2 mol HCN. 3 mol O2 2 mol HCN

3 L 02= 2 L HCN 3 L 02 2 LHCN

Calculate volume of oxygen gas (in L) required to react completely to produce 550 L of hydrogen cyanide

550 L HCNXHCN 3 L 02 2 L HON=825 L O

Hence, the volume of oxygen gas required to react completely to produce 550 L of hydrogen cyanide is 825 L.

B)

2CH49) + 2NH3(g) + 302(g) + 2HCN (9) + 6H2O(9)

2HCN + 6H2O

Under given conditions of temperature pressure, the ratio of number of moles of two gases (as per reaction stoichiometry) is equal to the ratio of the volumes of two gases.

6 mol H2O = 2 mol HCN. 6 mol H2O 2 mol HCN

6 L H20 = 2 L HCN 6 L H2O 2 LHCN

Calculate volume of water vapour (in L) required to react completely to produce 255 L of hydrogen cyanide

6 L H20 255 L HCN X- 2 L HCN = 765 L HO

Hence, the volume of water vapour required to react completely to produce 255 L of hydrogen cyanide is 765 L.