Hydrogen cyanide, HCN, can be made by a two-step process. First, ammonia is reacted with O2 to give nitric oxide, NO.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
Then nitric oxide is reacted with methane, CH4.
2 NO(g) + 2 CH4(g) → 2 HCN(g) + 2 H2O(g) + H2(g)
When 25.6 g of ammonia and 25.1 g of methane are used, how many grams of hydrogen cyanide can be produced?
I need help with this problem, and it's due tomorrow.
To find the number of grams of hydrogen cyanide (HCN) that can be produced, we need to determine the limiting reactant between ammonia (NH3) and methane (CH4). The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
First, we need to calculate the number of moles for each reactant using their respective molar masses:
Molar mass of NH3 = 17.03 g/mol
Molar mass of CH4 = 16.04 g/mol
Number of moles of NH3 = mass of NH3 / molar mass of NH3
= 25.6 g / 17.03 g/mol
Number of moles of CH4 = mass of CH4 / molar mass of CH4
= 25.1 g / 16.04 g/mol
Next, we need to calculate the stoichiometric ratios of the reactants based on the balanced chemical equation:
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
2 NO(g) + 2 CH4(g) → 2 HCN(g) + 2 H2O(g) + H2(g)
From the balanced equation, we can see that it takes 4 moles of NH3 to produce 2 moles of HCN.
Now, we can determine the limiting reactant. This can be done by comparing the stoichiometric ratios of NH3 and CH4 to the ratio of NH3 to HCN:
Ratio of NH3 to CH4 = 4/2 = 2 moles of NH3 per 1 mole of CH4
Ratio of NH3 to HCN = 4/2 = 2 moles of NH3 per 1 mole of HCN
Since the ratios are the same, both NH3 and CH4 are in the correct stoichiometric ratio. Therefore, both reactants are reacting in a 1:1 ratio.
Since we have equal amounts of NH3 and CH4, we can see that NH3 is the limiting reactant because it requires 4 moles to produce 2 moles of HCN, while CH4 only produces 1 mole of HCN.
Now, let's calculate the mass of HCN produced:
Number of moles of HCN = moles of NH3 (limiting reactant)
Mass of HCN = number of moles of HCN × molar mass of HCN
Molar mass of HCN = 27.03 g/mol
Number of moles of HCN = moles of NH3 (limiting reactant)
= (25.6 g / 17.03 g/mol) × (2 mol NH3 / 4 mol HCN)
Mass of HCN = number of moles of HCN × molar mass of HCN
= [(25.6 g / 17.03 g/mol) × (2 mol NH3 / 4 mol HCN)] × 27.03 g/mol
Simplifying the equation, we find:
Mass of HCN = (25.6 g × 2 × 27.03 g) / (17.03 g × 4)
= 51.22 g / 4
= 12.805 g
Therefore, approximately 12.805 grams of hydrogen cyanide can be produced when 25.6 g of ammonia and 25.1 g of methane are used.