3. 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? Show your work.
To find the answers to these questions, we will follow a step-by-step approach:
1. Balanced equation:
To balance the equation, we need to ensure that the number of atoms present in the reactants is equal to the number of atoms in the products. Here is the balanced equation:
CH4 + 2NH3 + 2O2 → HCN + 3H2O
2. Limiting reagent:
The limiting reagent is the reactant that is completely consumed in a chemical reaction, thus limiting the amount of product formed. To determine the limiting reagent, we compare the number of moles of each reagent to the stoichiometry of the balanced equation.
First, we need to convert the given masses of methane and ammonia to moles by using their molar masses:
- Methane (CH4): Molar mass = 12.01 g/mol + (1.01 g/mol * 4) = 16.05 g/mol
Moles of methane = 8 g / 16.05 g/mol ≈ 0.498 mol
- Ammonia (NH3): Molar mass = 14.01 g/mol + (1.01 g/mol * 3) = 17.04 g/mol
Moles of ammonia = 10 g / 17.04 g/mol ≈ 0.587 mol
Next, we compare the moles of each reactant to their respective coefficients in the balanced equation:
- For methane(CH4), the coefficient is 1.
- For ammonia(NH3), the coefficient is 2.
We can see that methane is in excess because it has fewer moles (0.498 mol) compared to ammonia (0.587 mol). Therefore, ammonia is the limiting reagent.
3. Grams of hydrogen cyanide formed:
To find the mass of hydrogen cyanide (HCN) formed, we need to use the stoichiometry from the balanced equation.
From the balanced equation, we see that the mole ratio of NH3 to HCN is 2:1.
So, for every 2 moles of NH3, we will get 1 mole of HCN.
Since ammonia is the limiting reagent, we use its moles to calculate the moles of HCN formed:
Moles of HCN = 0.587 mol × (1 mol HCN / 2 mol NH3) = 0.2935 mol
Now, we can calculate the mass of HCN by multiplying the moles with its molar mass:
Molar mass of HCN = 1.01 g/mol + 12.01 g/mol + 14.01 g/mol = 27.03 g/mol
Mass of HCN = 0.2935 mol × 27.03 g/mol ≈ 7.94 g
Therefore, approximately 7.94 grams of hydrogen cyanide will be formed.
To answer your questions:
1. Balanced Equation:
The balanced equation for the reaction between methane (CH4), ammonia (NH3), and oxygen (O2) to form hydrogen cyanide (HCN) and water (H2O) is as follows:
CH4 + 2NH3 + 3O2 → HCN + 3H2O
2. Limiting Reagent:
To determine the limiting reagent, we need to compare the number of moles of each reactant with their stoichiometric coefficients in the balanced equation.
Given:
Mass of methane (CH4) = 8 g
Mass of ammonia (NH3) = 10 g
First, let's convert the mass of each reactant to moles using their respective molar masses:
Molar mass of CH4 = 12.01 g/mol (C) + 4(1.008 g/mol) (H) = 16.04 g/mol
Molar mass of NH3 = 14.01 g/mol (N) + 3(1.008 g/mol) (H) = 17.03 g/mol
Number of moles of CH4 = 8 g / 16.04 g/mol = 0.499 mol
Number of moles of NH3 = 10 g / 17.03 g/mol = 0.586 mol
Next, compare the mole ratio of CH4 to NH3 in the balanced equation:
From the balanced equation: CH4 + 2NH3 + 3O2 → HCN + 3H2O
1 mol CH4 reacts with 2 mol NH3
By comparing the moles, we can see that the mole ratio of CH4 to NH3 is 1:2. Since we have fewer moles of CH4 (0.499 mol) compared to moles of NH3 (0.586 mol), CH4 is the limiting reagent.
3. Grams of Hydrogen Cyanide (HCN) formed:
To find the grams of HCN formed, we need to use the stoichiometry in the balanced equation.
From the balanced equation: CH4 + 2NH3 + 3O2 → HCN + 3H2O
1 mol CH4 reacts with 1 mol HCN
Using the moles of CH4 (0.499 mol), we can calculate the moles of HCN formed:
Moles of HCN = 0.499 mol
Next, we convert the moles of HCN to grams using its molar mass:
Molar mass of HCN = 1.01 g/mol (H) + 12.01 g/mol (C) + 14.01 g/mol (N) = 27.03 g/mol
Grams of HCN formed = Moles of HCN × Molar mass of HCN
= 0.499 mol × 27.03 g/mol
= 13.48 g
Therefore, approximately 13.48 grams of hydrogen cyanide (HCN) will be formed.
4CH4 + 4NH3 + 6O2 ==> 4HCN + 12H2O
You will never balance by trial and error; here is a good site that will balance redox equations.
http://www.chemteam.info/Redox/Redox.html
mols NH3 = 10g/17 = about 0.6 but you need to do it more accurately.
mols CH4 8/16 = about 0.50
Convert mol NH3 and mols CH4 to mols HCN.
0.6mol NH3 x (4 mols HCN/4 mol NH3) = ?
0.0.5 mol CH4 x (4 mol HCN/4 mol CH4) = ?
You have two values for mols HCN and both can't be right. In limiting reagent problems the SMALLER number is ALWAYS the correct value.
Convert that value to g. g = mols x molar mass.