Separation of components of crude oil
The grams of oxygen that are required to produce 1 mole of H₂O is 16 g ( answer B)
<u><em> calculation</em></u>
2 CH₄ + 2NH₃ +3 O₂ → 2HCN + 6H₂O
step 1: use the mole ratio to find moles of O₂
from equation above the mole ratio of O₂: H₂O is 3:6 therefore the moles of O₂ = 1 mole x3/6 =0.5 moles
step 2: find mass of O₂
mass= moles x molar mass
from periodic table the molar mass of O₂ = 16 x2= 32 g/mol
mass O₂ = 0.5 moles x 32 g/mol = 16 g (answer B)
I believe d is the correct answer
A) in pure water :
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0
change +X +X
Equ X X
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X* X
∴X = √2.1 x 10*-10
∴X = 1.4 x 10^-5
∴ the solubility = X = 1.4 X 10^-5
B) In 1.6 x 10^-3 m Na2CrO4
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0.0016
Change +X +X
Equ X X+0.0016
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X*(X+0.0016) by solving for X
∴ X = 1.3 x 10^-7
∴ solubility =X = 1.3 x 10^-7
