First convert grams to moles
using molar mass of butane that is 58.1 g
3.50g C4H10 x (1 mol
C4H10)/(58.1g C4H10) = 0.06024 mol C4H10 <span>
<span>Now convert moles to molecules by using Avogadro’s number
0.06024 mol C4H10 x (6.022x10^23 molecules C4H10)/(1 mol
C4H10) = 3.627x10^22 molecules C4H10
And there are 4 carbon atoms in 1 molecule of butane, so use
the following ratio:
3.627 x 10^22 molecules C4H10 x (4 atoms C)/(1 molecule
C4H10)
<span>= 1.45 x 10^23 atoms of carbon are present</span></span></span>
Woah, that doesn't look like high school. That's gotta be college level.
Answer:
Kc = 4.76
Explanation:
To find the concentrations of CO and H₂ at equilibrium, you have to set up an ICE (Initial, Change, Equilibrium) table.
CO (g) + 2H₂(g) ⇌ CH₃OH (g)
I 0.32 M 0.53 M 0
C -x -2x +x
E 0.32-x 0.53-2x 0.16 M
Since you know the concentration of CH₃OH at equilibrium, it would be equal to x since 0 + x = 0.16. So,
[CH₃OH] = 0.16 M
[CO] = 0.32 - 0. 16 = 0.16 M
[H₂] = 0.53 - 2(0.16) = 0.21 M
Now that you have all the concentrations at equilibrium, you can calculate the equilibrium constant.
Kc = products ÷ reactants
= [CH₃OH] ÷ [CO][H₂]²
= 0.16 ÷ (0.16)(0.21)
Kc = 4.76
The equilibrium constant at this temperature is 4.76.
Hope this helps.