It would be C
2 kg x 1000 g/kg x 1mol/18.02 x 6.03 kj/mol = 669kj
Oxidation reaction
In ---> In³⁺ + 3e ---1)
reduction reaction
Cd²⁺ + 2e ---> Cd ---2)
when balancing the reactions, electrons have to be balanced. to balance the electrons multiple 1st reaction by 2 and 2nd reaction by 3
1) x 2
2) x 3
2In ---> 2In³⁺ + 6e
3Cd²⁺ + 6e ---> 3Cd
add the 2 equations to obtain the overall reaction
2In + 3Cd²⁺ ---> 2In³⁺ + 3Cd
NH₃:
N = 8*10²²
NA = 6.02*10²³
n = N/NA = 8*10²²/6.02*10²³ ≈ 1.33*10⁻¹=0.133mol
O₂:
N=7*10²²
NA = 6.02*10²³
n = N/NA = 7*10²²/6.02*10²³ = 1.16*10⁻¹=0.116mol
4NH₃ <span>+ 3O</span>₂ ⇒<span> 2N</span>₂<span> + 6H</span>₂<span>O
</span>4mol : 3mol : 2mol
0.133mol : 0.116mol : 0,0665mol
limiting reactant
N₂:
n = 0.0665mol
M = 28g/mol
m = n*M = 0.0665mol*28g/mol = <u>1,862g</u>
Answer:
let's go to the beach or u will relax lol
