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?

See your other post above.

thank you

The balanced equation for the reaction is:

CH4 + 2NH3 + 3O2 -> HCN + 3H2O

To determine which reagent is limiting, we need to calculate the amount of hydrogen cyanide that can be formed from each reactant.

First, let's calculate the amount of hydrogen cyanide that can be formed from methane. The molar mass of methane (CH4) is 16 g/mol. So, the number of moles of methane is:

8 g / 16 g/mol = 0.5 mol

According to the balanced equation, the stoichiometric ratio between methane and hydrogen cyanide is 1:1. Therefore, 0.5 mol of methane can produce 0.5 mol of hydrogen cyanide.

Next, let's calculate the amount of hydrogen cyanide that can be formed from ammonia. The molar mass of ammonia (NH3) is 17 g/mol. So, the number of moles of ammonia is:

10 g / 17 g/mol = 0.588 mol

According to the balanced equation, the stoichiometric ratio between ammonia and hydrogen cyanide is 2:1. Therefore, 0.588 mol of ammonia can produce (0.588 mol / 2) = 0.294 mol of hydrogen cyanide.

Comparing the amounts of hydrogen cyanide that can be formed from methane (0.5 mol) and ammonia (0.294 mol), we can see that ammonia is the limiting reagent. This is because it produces a lesser amount of hydrogen cyanide compared to methane.

To calculate the mass of hydrogen cyanide that will be formed, we need to multiply the moles of hydrogen cyanide by its molar mass. The molar mass of hydrogen cyanide (HCN) is 27 g/mol. So, the mass of hydrogen cyanide formed is:

0.294 mol * 27 g/mol = 7.958 g

Therefore, 7.958 grams of hydrogen cyanide will be formed in this reaction.

To answer the questions, let's go through each step one by one.

1. Balanced Equation:
The balanced equation for the reaction is:
CH4 + 2NH3 + 2O2 → HCN + 3H2O

2. Limiting Reagent:
To determine the limiting reagent, we compare the number of moles of each reactant to their stoichiometric coefficients in the balanced equation.

For methane (CH4), we have 8 g. To find the number of moles, we use the formula:
number of moles = mass / molar mass

The molar mass of methane (CH4) is calculated as:
(1x12.01 g/mol) + (4x1.008 g/mol) = 16.04 g/mol

Thus, the number of moles of methane is:
moles of CH4 = 8 g / 16.04 g/mol = 0.498 mol

For ammonia (NH3), we have 10 g. The molar mass of ammonia (NH3) is:
(1x14.01 g/mol) + (3x1.008 g/mol) = 17.03 g/mol

And the number of moles of ammonia is:
moles of NH3 = 10 g / 17.03 g/mol = 0.587 mol

Now, let's compare the number of moles of each reactant with their stoichiometric coefficients in the balanced equation:

For methane, the coefficient is 1.
For ammonia, the coefficient is 2.

moles of CH4 / coef. of CH4 = 0.498 mol / 1 = 0.498 mol
moles of NH3 / coef. of NH3 = 0.587 mol / 2 = 0.2935 mol

We can see that the ratio of moles for methane to ammonia is approximately 2:1. This implies that we require twice the number of moles of ammonia compared to methane for the reaction. However, in this case, we have fewer moles of ammonia than methane. Therefore, ammonia is the limiting reagent.

3. Calculating the Mass of Hydrogen Cyanide:
Now, let's find the number of moles of hydrogen cyanide (HCN) formed using the limiting reagent (ammonia).

From the balanced equation, we see that the stoichiometric coefficient of HCN is 1.

moles of HCN = moles of limiting reagent (NH3) = 0.2935 mol

To calculate the mass of HCN formed, we use the formula:
mass = molar mass × number of moles

The molar mass of HCN is:
(1x1.008 g/mol) + (1x12.01 g/mol) + (1x14.01 g/mol) = 27.03 g/mol

Therefore, the mass of HCN formed is:
mass of HCN = 27.03 g/mol × 0.2935 mol = 7.93 g

Hence, approximately 7.93 grams of hydrogen cyanide will be formed.