The reaction for the industrial production of hydrogen cyanide is given below.

2CH4 + 2NH3 + 3O2 --> 2HCN + 6H2O

What mass of NH3 is required for the synthesis of 1.5 × 105 kg of HCN if the reaction has a 97.5% yield?

To determine the mass of NH3 required for the synthesis of 1.5 × 105 kg of HCN, we can use the stoichiometry of the reaction.

The balanced equation for the reaction is: 2CH4 + 2NH3 + 3O2 → 2HCN + 6H2O

From the equation, we can see that 2 moles of NH3 are required to produce 2 moles of HCN.

To find the number of moles of HCN produced, we can use the given mass and molar mass of HCN.

Molar mass of HCN:
H (1.00784 g/mol) + C (12.0107 g/mol) + N (14.00674 g/mol) = 27.0243 g/mol

Number of moles of HCN = mass of HCN / molar mass of HCN
Number of moles of HCN = 1.5 × 105 kg / (27.0243 g/mol) / (1000 g/kg)
Number of moles of HCN = 5555.55 mol

Since the reaction has a 97.5% yield, this means that only 97.5% of the HCN will be produced.

Actual moles of HCN produced = 97.5% × 5555.55 mol
Actual moles of HCN produced = 5409.81 mol

From the stoichiometry of the reaction, we can see that 2 moles of NH3 are required to produce 2 moles of HCN.

Therefore, the moles of NH3 required = moles of HCN produced = 5409.81 mol

Finally, we can calculate the mass of NH3 required by using the molar mass of NH3:

Molar mass of NH3:
N (14.00674 g/mol) + H (1.00784 g/mol) × 3 = 17.03052 g/mol

Mass of NH3 required = moles of NH3 required × molar mass of NH3
Mass of NH3 required = 5409.81 mol × 17.03052 g/mol
Mass of NH3 required = 92,147.73 g = 92.15 kg

Therefore, approximately 92.15 kg of NH3 is required for the synthesis of 1.5 × 105 kg of HCN with a 97.5% yield.

To determine the mass of NH3 required for the synthesis of 1.5 × 10^5 kg of HCN, we need to use stoichiometry and take into account the reaction yield.

First, let's balance the equation to ensure that the number of atoms for each element is equal on both sides of the equation:

2CH4 + 2NH3 + 3O2 → 2HCN + 6H2O

The stoichiometry of the balanced equation tells us that 2 moles of NH3 are required to produce 2 moles of HCN. Therefore, the molar ratio is 1:1 between NH3 and HCN.

Given that the reaction yield is 97.5%, this means that only 97.5% of the theoretical maximum amount of HCN is obtained. Therefore, we need to adjust the amount of the desired product (1.5 × 10^5 kg) accordingly.

First, we calculate the theoretical maximum amount of HCN produced:

Mass of HCN = 1.5 × 10^5 kg ÷ 0.975 = 1.538 × 10^5 kg

Now, since the molar ratio of NH3 to HCN is 1:1, we can directly relate the mass of NH3 to the mass of HCN. The molar mass of NH3 is 17.03 g/mol, and the molar mass of HCN is 27.03 g/mol.

Let's calculate the amount of NH3 needed:

Moles of HCN = Mass of HCN / Molar mass of HCN
Moles of HCN = (1.538 × 10^5 kg) / (27.03 g/mol)
Moles of HCN = 5.69 × 10^6 mol

Since the molar ratio between NH3 and HCN is 1:1, the moles of NH3 needed are also 5.69 × 10^6 mol.

Finally, let's convert the moles of NH3 to mass:

Mass of NH3 = Moles of NH3 × Molar mass of NH3
Mass of NH3 = (5.69 × 10^6 mol) × (17.03 g/mol)
Mass of NH3 = 9.68 × 10^7 g

Therefore, approximately 9.68 × 10^7 grams (or 96.8 metric tons) of NH3 are required for the synthesis of 1.5 × 10^5 kg of HCN, considering a 97.5% yield.