Hydrogen cyanide is used in the production of cyanimid fertilisers. It is produced by the
following reaction:
2CH4 + 2NH3 + 3O2 2HCN + 6H2O
(a) How much hydrogen cyanide can be produced starting with 100 kg of each of the
reactants?
(b) Which is the limiting reagent?
14,188 mol HCN
Oxygen gas is the limiting reactant
To find out how much hydrogen cyanide can be produced starting with 100 kg of each of the reactants, we need to follow these steps:
Step 1: Convert the given mass of reactants (in this case, 100 kg) into moles.
Step 2: Use the balanced chemical equation to determine the stoichiometric ratio between the reactants and products.
Step 3: Identify the limiting reagent by comparing the moles of each reactant.
Let's calculate the amount of hydrogen cyanide produced:
(a) How much hydrogen cyanide can be produced starting with 100 kg of each of the reactants?
Step 1: Convert 100 kg of each reactant into moles.
- The molar mass of CH4 (methane) is 16 g/mol.
100 kg CH4 = (100,000 g CH4) / (16 g/mol) = 6250 mol CH4
- The molar mass of NH3 (ammonia) is 17 g/mol.
100 kg NH3 = (100,000 g NH3) / (17 g/mol) = 5882 mol NH3
- The molar mass of O2 (oxygen) is 32 g/mol.
100 kg O2 = (100,000 g O2) / (32 g/mol) = 3125 mol O2
Step 2: Use the stoichiometric coefficients from the balanced equation to calculate the moles of HCN produced.
According to the balanced equation:
2 CH4 + 2 NH3 + 3 O2 → 2 HCN + 6 H2O
The stoichiometric ratio between CH4, NH3, O2, and HCN is 2:2:3:2. This means that for every 2 moles of CH4, 2 moles of NH3, and 3 moles of O2, 2 moles of HCN are produced.
From the calculations in Step 1, we have:
- Moles of CH4 = 6250 mol
- Moles of NH3 = 5882 mol
- Moles of O2 = 3125 mol
But in order to produce HCN, we need the same number of moles of CH4 and NH3, and 1.5 times the number of moles of O2.
Step 3: Calculate the limiting reagent by comparing the moles of each reactant.
CH4 and NH3 have the same number of moles, which means they are not the limiting reagents.
To find the limiting reagent, we compare the ratio of O2 to HCN.
Moles of HCN produced = (3125 mol O2) / (3 mol O2 per 2 mol HCN) = 2083.33 mol HCN
Since the stoichiometric ratio between O2 and HCN is 3:2, we have produced excess HCN.
So, the limiting reagent is O2.
(b) The limiting reagent is O2 because it can only produce 2083.33 moles of HCN, while CH4 and NH3 can produce more.