2. Methane (CH4), ammonia (NH3), and oxygen (O2) can react to form hydrogen cyanide (HCN) and water according to this equation:
CH4 + NH3 + O2 HCN + H2O
You have 8 g of methane and 10 g of ammonia in excess oxygen. Answer the following questions:
• What is the balanced equation for this reaction?
• Which reagent is limiting? Explain why.
• How many grams of hydrogen cyanide will be formed?
2CH4 + 2NH3 +3O2 Arrow 2HCN + 6H2O
CH4 is going to be used up making it the limited reagent
13.5 g CH4
10gNH3(1 mole NH3)(2 mole HCN)
(14.01g NH3)(2 mole NH3)
=.714 mole HCN
Do the same for Ammonia and you get .499 mole HCN or .50
To answer the questions, let's start by balancing the equation:
CH4 + 2NH3 + 2O2 → HCN + 3H2O
Now, let's determine which reagent is limiting. The limiting reagent is the one that is completely consumed, thus determining the maximum amount of product that can be formed.
First, let's calculate the moles of methane (CH4). The molar mass of methane is 16.04 g/mol, so we have:
8 g CH4 × (1 mol CH4 / 16.04 g CH4) = 0.498 mol CH4
Now, let's calculate the moles of ammonia (NH3). The molar mass of ammonia is 17.03 g/mol, so we have:
10 g NH3 × (1 mol NH3 / 17.03 g NH3) = 0.587 mol NH3
Next, let's compare the moles of methane and ammonia to their coefficients in the balanced equation. The coefficient of methane is 1, while the coefficient of ammonia is 2. This means that we need twice as many moles of ammonia as methane for the reaction to be balanced.
To find the limiting reagent, we can compare the moles of the two reagents. Since we need twice as many moles of ammonia, but we have fewer moles of methane, methane is the limiting reagent.
Now, let's calculate the grams of hydrogen cyanide (HCN) that will be formed. According to the balanced equation, the molar ratio between methane and hydrogen cyanide is 1:1. The molar mass of hydrogen cyanide is 27.03 g/mol.
Therefore, the number of moles of hydrogen cyanide formed is equal to the number of moles of methane:
0.498 mol HCN
Finally, let's convert moles to grams using the molar mass of hydrogen cyanide:
0.498 mol HCN × (27.03 g HCN / 1 mol HCN) = 13.45 g HCN
So, 13.45 grams of hydrogen cyanide will be formed.
To determine the balanced equation for this reaction, we need to ensure that the number of atoms of each element is equal on both sides of the equation.
The balanced equation is:
CH4 + 2NH3 + 3O2 → HCN + 3H2O
To determine which reagent is limiting, we need to compare the number of moles of the reactants with the stoichiometric ratio in the balanced equation. The reagent that produces the least amount of product is the limiting reagent.
First, we need to calculate the number of moles for each reactant:
Molar mass of CH4 = 16.04 g/mol
Number of moles of CH4 = mass / molar mass = 8 g / 16.04 g/mol = 0.499 mol
Molar mass of NH3 = 17.03 g/mol
Number of moles of NH3 = mass / molar mass = 10 g / 17.03 g/mol = 0.587 mol
Next, we can compare the moles of CH4 and NH3 with their stoichiometric ratio in the balanced equation:
CH4 : NH3 : O2 = 1 : 2 : 3
From the balanced equation, we can see that for every 1 mole of CH4, we need 2 moles of NH3 and 3 moles of O2.
Let's compare CH4 with NH3 first:
0.499 mol CH4 : 0.587 mol NH3 ≈ 1 : 1.17
Since the ratio is approximately 1:1, we can say that both CH4 and NH3 are in equal stoichiometry.
Now, let's compare CH4 with O2:
0.499 mol CH4 : (0.499 mol CH4 × 3 mol O2 / 1 mol CH4) = 1.497 mol O2
Since the ratio is less than 1:3, CH4 is the limiting reagent.
To find the amount of HCN formed, we need to use the stoichiometric ratio between CH4 and HCN from the balanced equation:
1 mol CH4 : 1 mol HCN
Since CH4 is the limiting reagent, the amount of HCN formed will be equal to the amount of CH4 used.
Number of moles of HCN formed = number of moles of CH4 = 0.499 mol
To convert the moles of HCN to grams, we can use the molar mass of HCN:
Molar mass of HCN = 27.03 g/mol (1 H + 12.01 C + 14.01 N)
Mass of HCN formed = number of moles of HCN × molar mass of HCN
Mass of HCN formed = 0.499 mol × 27.03 g/mol ≈ 13.496 g
Therefore, approximately 13.496 grams of hydrogen cyanide (HCN) will be formed.