Answer:
For every 4 moles of NO created, 6 moles of H2O are created so the ratio is 4:6
Explanation:
You just need to balance the equation.
NH3 + O2 -> NO + H2O
1. I started with hydrogen; there's 3 on the left and 2 on the right. Multiply them together to find a number they both go into (3×2=6, but in this case 6 hydrogen on each side does not work so I doubled it so there is 12 hydrogen on each side).
This will bring you to this:
4NH3 + O2 -> NO + 6H2O
2. Now get equal amounts of nitrogen on each side. There's 4 nitrogen on the left side, and 1 on the right. Multiply the right by 4. Then you will have this:
4NH3 + O2 -> 4NO + 6H2O
3. Last thing you need to do is have the same amount of oxygen on both sides. On the left you have 2 and on the right you have 10. Get the left to 10 by multiplying it by 5.
Balanced: 4NH3 + 5O2 -> 4NO + 6H2O
In word form, for every reaction between 4 moles of ammonia and 5 moles of oxygen, 4 moles of nitric oxide and 6 moles of water will be created.
I hope this helps!
Pretty much, if I were going to separate small solid particles, I could use like a piece of paper. I used some type of piece of paper when I was trying to separate some particles during science.
The equilibrium constant, Kc=0.026
<h3>Further explanation</h3>
Given
1.72 moles of NOCI
1.16 moles of NOCI remained
2.50 L reaction chamber
Reaction
2NOCI(g) = 2NO(g) + Cl2(g).
Required
the equilibrium constant, Kc
Solution
ICE method
2NOCI(g) = 2NO(g) + Cl2(g).
I 1.72
C 0.56 0.56 0.28
E 1.16 0.56 0.28
Molarity at equilibrium :
NOCl :
NO :
Cl2 :
![\tt Kc=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\\\\Kc=\dfrac{0.224^2\times 0.112}{0.464^2}=0.026](https://tex.z-dn.net/?f=%5Ctt%20Kc%3D%5Cdfrac%7B%5BNO%5D%5E2%5BCl_2%5D%7D%7B%5BNOCl%5D%5E2%7D%5C%5C%5C%5CKc%3D%5Cdfrac%7B0.224%5E2%5Ctimes%200.112%7D%7B0.464%5E2%7D%3D0.026)
