Partial mass of C, 12*2 = 24
Partial mass of H, 1*5 = 5
Partial mass of N, 14
Partial mass of O, 16
Now, total molar mass would be 24+5+14+16 = 59
mass % of carbon = 24 * 100 / 59 = 40.67%
Use stoich
3.01 x 10^23 Au* 1m / 6.022 x 10^23
= .499m Au
Explanation:
2 C(s) + 3 H2(g) = C2H6(g)
We need to use the following formula
Δ
![G= -nF E_{cell}](https://tex.z-dn.net/?f=G%3D%20-nF%20E_%7Bcell%7D%20)
![E_{cell}= 0.24V](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D%200.24V)
n= 4 moles
F= constant= 96500C/mol
let's plug in the values.
ΔG= -(4)(96500)(0.24)=
-92640 J or -92.6 kJ