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?
THANK YOU SOOOO MUCH!!!!! I WAS SO CONFUSED!!!
2CH4 + 2NH3 + 3O2 2HCN + 6H2O
mols CH4 = grams/molar mass CH4 =?
mols NH3 = grams/molar mass NH3 = ?
a. Convert mols CH4 to mols HCN using the coefficients in the balanced equation.
b. Convert mols NH3 the same way.
c. It is likely that the answers from a and b will not be the same which means one of them is wrong.
d. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent.
Using the smaller value convert mols to grams. g = mols HCN x molar mass HCN
To answer your questions step-by-step:
Step 1: Balancing the equation:
The balanced equation for the reaction is as follows:
CH4 + 2NH3 + 3O2 -> HCN + 3H2O
Step 2: Determining the limiting reagent:
To determine the limiting reagent, we need to compare the number of moles of each reagent to the balanced equation. The reagent that produces the fewest moles is the limiting reagent.
For methane (CH4):
Molar mass of CH4:
C = 12.01 g/mol
H = 1.01 g/mol (4 hydrogens x 1.01 g/mol = 4.04 g/mol)
Total molar mass = 12.01 g/mol + 4.04 g/mol = 16.05 g/mol
Moles of CH4 = Mass / Molar mass
Moles of CH4 = 8 g / 16.05 g/mol ≈ 0.498 mol
For ammonia (NH3):
Molar mass of NH3:
N = 14.01 g/mol
H = 1.01 g/mol (3 hydrogens x 1.01 g/mol = 3.03 g/mol)
Total molar mass = 14.01 g/mol + 3.03 g/mol = 17.04 g/mol
Moles of NH3 = Mass / Molar mass
Moles of NH3 = 10 g / 17.04 g/mol ≈ 0.587 mol
For oxygen (O2):
Molar mass of O2 = 32.00 g/mol
Since oxygen is in excess, we do not need to calculate the moles.
Comparing the moles of reactants, we see that methane has fewer moles (0.498 mol) compared to ammonia (0.587 mol). Therefore, methane is the limiting reagent.
Step 3: Calculating the grams of hydrogen cyanide:
From the balanced equation, we can see that 1 mole of CH4 reacts to form 1 mole of HCN.
Moles of HCN = Moles of CH4
Moles of HCN = 0.498 mol
Molar mass of HCN:
H = 1.01 g/mol
C = 12.01 g/mol
N = 14.01 g/mol
Total molar mass = 1.01 g/mol + 12.01 g/mol + 14.01 g/mol = 27.03 g/mol
Grams of HCN = Moles of HCN * Molar mass of HCN
Grams of HCN = 0.498 mol * 27.03 g/mol ≈ 13.46 g
Therefore, approximately 13.46 grams of hydrogen cyanide will be formed.
To answer these questions, we can follow a step-by-step approach:
1. Balancing the equation:
The balanced equation for the reaction is:
CH4 + 2NH3 + 3O2 → HCN + 3H2O
2. Determining the limiting reagent:
To determine the limiting reagent, we need to compare the amounts of reactants we have with the stoichiometry of the balanced equation.
- Start with the amount of methane (CH4): 8 g.
- Next, calculate the molar mass of methane (CH4):
Molar mass of CH4 = 12.01 g/mol + 4(1.01 g/mol) = 16.05 g/mol
- Use the molar mass to convert grams of methane to moles:
Moles of CH4 = mass (g) / molar mass (g/mol)
= 8 g / 16.05 g/mol
≈ 0.498 mol (rounded to three decimal places)
- Now, consider the amount of ammonia (NH3): 10 g.
- Calculate the molar mass of ammonia (NH3):
Molar mass of NH3 = 14.01 g/mol + 3(1.01 g/mol) = 17.03 g/mol
- Convert grams of ammonia to moles:
Moles of NH3 = mass (g) / molar mass (g/mol)
= 10 g / 17.03 g/mol
≈ 0.587 mol (rounded to three decimal places)
- Finally, determine the limiting reagent by comparing the ratios of the coefficients:
Moles of CH4: Moles of NH3: Moles of O2 = 0.498 mol : 0.587 mol : Excess
- From the balanced equation, we can see that the ratio of CH4 to NH3 is 1:2. So, for every 1 mole of CH4, we need 2 moles of NH3 to react completely.
- Since the ratio of the moles of CH4 to NH3 is less than 1:2, CH4 is the limiting reagent in this reaction.
3. Calculating the grams of hydrogen cyanide (HCN) formed:
To determine the grams of HCN formed, we will use the stoichiometry in the balanced equation.
- From the balanced equation, the mole ratio of CH4 to HCN is 1:1.
- Since CH4 is the limiting reagent and we have 0.498 mol of CH4, we will have the same amount of moles of HCN produced.
- Calculate the molar mass of hydrogen cyanide (HCN):
Molar mass of HCN = 1.01 g/mol + 12.01 g/mol + 14.01 g/mol = 27.03 g/mol
- Convert moles of HCN to grams:
Grams of HCN = moles of HCN × molar mass of HCN
= 0.498 mol × 27.03 g/mol
≈ 13.47 g (rounded to two decimal places)
Therefore, approximately 13.47 grams of hydrogen cyanide (HCN) will be formed in this reaction.