The answer is A :) homogeneous!
<span>Fe2O3 + 3CO --> 2Fe + 3CO2
</span><span>
m(Fe2O3)=213 g
m(CO)=140 g
</span>_______________
<span>n(Fe2O3)=?
m(Fe)=?
n(Fe2O3)=?
n(CO)=?
n(CO2)=?
</span>
<span>n(Fe2O3)=m(Fe2O3) / M(Fe2O3)
n(Fe2O3)= 213 g / 159,7 gmol-1 = 1,33 mol
</span>
<span>n(CO)= m(CO) / M(CO)
n(CO)= 140 g / 28,01 gmol-1 = 4,99 mol</span>
Answer:
Aluminium (aluminum in American and Canadian English) is a chemical element with the symbol Al and atomic number 13. It is a silvery-white, soft, non-magnetic and ductile metal in the boron group. ... Aluminium is remarkable for its low density and its ability to resist corrosion through the phenomenon of passivation.
Explanation:
i did aluminum chemistry
<em>ANSWER - 6 MOLES OF </em><em>IRON</em>
Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(1)
One moles of Fe2O3 forming 2 moles of Fe
3 moles of Fe2O3 will form 2×3 = 6 moles of iron
Answer : The concentration of 
and 
at equilibrium is, 0.0158 M and 0.00302 M respectively.
Explanation :
First we have to calculate the concentration of 
Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:
Initial conc. 0.0163 0.00415 0.00276
At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)
As we are given:
Concentration of 
at equilibrium = 0.00467 M
That means,
(0.00415+x) = 0.00467
x = 0.00026 M
Concentration of at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M
Concentration of at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M
