Answer:
32
Explanation:
because the number of heliem is your aswer
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
I’m just commenting rn to get points
Answer:
3NaOH (aq) + Fe(NO₃)₃ (aq) → Fe(OH)₃ (s) + 3NaNO₃ (aq)
Explanation:
Step 1: RxN
NaOH (aq) + Fe(NO₃)₃ (aq) → Fe(OH)₃ (s) + NaNO₃ (aq)
Step 2: Balance RxN
We need 3 OH's on both sides.
We also need 3 NO₃'s on both side.
- This will make it so we also need 3 Na's on both side
3NaOH (aq) + Fe(NO₃)₃ (aq) → Fe(OH)₃ (s) + 3NaNO₃ (aq)
