Answer:
34.3 g NH3
Explanation:
M(H2) = 2*1 = 2 g/mol
M(N2) = 2*14 = 28 g/mol
M(NH3) = 14 + 3*1 = 17 g/mol
23.6 g H2* 1 mol/2 g = 11.8 mol H2
28.3 g N2 * 1 mol/28 g = 1.01 mol N2
3H2 + N2 ------> 2NH3
from reaction 3 mol 1 mol
given 11.8 mol 1.01 mol
We can see that H2 is given in excess, N2 is limiting reactant.
3H2 + N2 ------> 2NH3
from reaction 1 mol 2 mol
given 1.01 mol x
x = 2*1.01/1= 2.02 mol NH3
2.02 mol * 17g/1 mol ≈ 34.3 g NH3
Iron bromide. Iron bromide (FeBr3)
Reactants + Energy → Products
I guess this is the answer
You’re welcome ;)
Answer:
There are 0.5 mole in 20g of argon.
Explanation:
40 g of argon = 1mole
Then 20g of argon is,
→ 1/40 × 20
→ 0.5 mole
There are 2 unpaired electrons in sulphur orbital...
