ok, so sorry, buuuuut..... I need help with this desperately...
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? Show your work.

For your questions, these are the answers:
1. 2CH4 + 2NH3 + 302 -> 2HCN + 6H2O

2. Methane
2mole CH4 = 32g
react with 2 mole NH3 which is equal to 34g
so,
8g CH4 react with 34/32 x 8
= 8.5g NH3

3. 32g methane = 54g HCN (2 moles)
so,
8g CH4 = 54/32 x 8g HCN
=1.60 g HCN
answer: 1.60 g cyanide

Answer Link

joetheelite joetheelite · Answer 2 · 2019-12-19T17:17:15+01:00

Answer:

See explanation

Explanation:

First, let's write the equation again.

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

In order to balance this equation, we need to see both sides of the reaction (Reactants and products), and count the elements and atoms there. it's usually begin with the metals, then non metals, and finally hydrogen and oxygen.

Let's begin with the Carbon and nitrogen.

In reactant we have 1 C and product the same, so theorically speaking, we don't need to balance, same thing happen with N (1 and 1).

However, when we look at H, we can see we have 7 in reactants (4 and 3) and only 3 in products (1 and 2), so we need to balance them. As we have those atoms in different compounds, we need to put a number in both compound so the sum gives an even number. In this case if we put a 2 in CH4, we'll get 8 hydrogen and the other 3, will be 11.

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

To get an even number, let's put a 2 in NH3 too, the total is now 14.

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

Let get 14 in the products, to do that we simply put a 2 in HCN and 6 in H2O.

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

With that, we balance Hydrogen, and also Carbon and nitrogen were balanced too with this.

Now only the oxygen needs to be balanced. The 6 in H2O put the oxygen with 6, so we need to put a 3 in reactants, and with that, the equation is balanced:

2CH4 + 2NH3 + 3O2 -----> 2HCN + 6H2O

Now that we have the balanced equation, we can calculate the rest of the questions.

The limiting reagent is the reagent that it gets consumed first and completely in the reaction. To do that, we need to do stechiometry of the reagents, and the easier way to do that is with the number of moles.

moles are calculated:

n = m/MM

The molecular mass of CH4 is 16 g/mol and NH3 is 17 g/mol. According to the balanced equation, we have at least 2 moles of CH4 and 2 moles of NH3. The limiting reagent will be the reagent with the lower moles so:

nCH4 = 8/16 = 0.5 moles

nNH3 = 10/17 = 0.59 moles

We have more moles of NH3 than moles CH4, so the limiting reagent is CH4.

Now to get the mass of HCN, we need the moles. But we already know which is the limiting reagent and we know (according to the balanced reaction) that 2 moles of CH4 produces 2 moles HCN, so, the moles of CH4 are the same of HCN produced therefore:

moles CH4 = moles HCN = 0.5 moles

To get the mass we need the molecular mass of HCN which is 27 g/mol, so the mass:

m = 27 * 0.5

m = 13.5 g

This is the mass formed of HCN