Using the balanced reaction for hydrogen cyanide, if 3.84 moles of methane are reacted, how many moles of H2O will be consumed?

Question

anonymous · Answer

First of all, what's the balanced equation?

anonymous · Answer

Until you get the balanced equation, refer to this, "two moles of water are formed per one mole of methane consumed

anonymous · Answer

$$2NH _{3} + 3O _{2} + 2CH _{4} ---> 2HCN + 6H _{2}O$$

anonymous · Answer

actually the question first asks, how many moles of oxygen will be consumed, then how many moles of H2O will be produced

anonymous · Answer

Okay well that changes everything.

anonymous · Answer

I'm sorry

anonymous · Answer

No, it's just different steps. I wasn't complaining.

anonymous · Answer

For 2 moles of NH3, you get 3 moles of O2. So the ratio is :
NH3 : O2
2 : 3

And, you know that3.84 moles of NH3 are used.

So, let x be the number of moles of oxygen consumed. 

NH3 : O2
2 : 3
3.84 : x 

It's simple multiplication/division. x = (3.84 times 3 ) divided by 2
                                                        = 5.76 moles
Therefore, for 3.84 moles of NH3 consumed, 5.76 moles of O2 are consumed.

And now for the second part.

anonymous · Answer

This is the same as the first. Write the ratio of NH3 : H2O and let the moles of H2O be x. 

NH3 : H2O
2 : 6
3.84 : x
and then find x. 

x = (3.84 * 6) / 2
   = 11.52 moles. therefore, 11.52 moles of water is produced. 

I used the ratio of NH3 : H2O, but you could also use the ratio of O2 : H2O

anonymous · Answer

ok, so the first part of the question you are using the molecules on the left side of the equation, right?

anonymous · Answer

Yes

anonymous · Answer

I'm comparing what I know and I need to find out.

anonymous · Answer

OH ok, I was trying to figure it out using 6 for oxygen

anonymous · Answer

What do you mean using 6? You use whatever is infront of the molecule. That's the molar ratio.

anonymous · Answer

yeah I was getting ahead of myself I guess bc the next question is if 25.1g of NH3 are reacted how many g of hydrogen cyanide are produced

anonymous · Answer

thank you so much

anonymous · Answer

With this question, it's the same things. FInd the number of moles of HCN by comparing NH3 and HCN (the answer will be 3.84 moles). Then use the formule n = m/MM (moles = mass/Molar Mass)
If you rearrange the formula, you get m = n x MM
n is your moles, which is 3.84 and find your Molar Mass by using the periodic table. Then just sub in the numbers and you'll have your mass of HCN produced.