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
Answer:
Whats the hypothesis and the experiment?
Explanation:
I cant really help without context
They have low boiling points
Answer:
34.9 g of Zn(OH)₂ is the maximum mass that can be formed
Explanation:
Let's state the reaction:
ZnO(s) + H₂O(l) → Zn(OH)₂ (aq)
First of all, we need to determine the moles of each reactant and state the limiting:
28.6 g . 1mol /81.38 g = 0.351 moles of ZnO
9.54 g . 1mol /18 g = 0.53 moles of water
As ratio is 1:1, for 0.53 moles of water, we need 0.53 moles of ZnO, but we only have 0.351, so the limiting reactant is the ZnO.
Ratio with the product is also 1:1. From 0.351 moles of oxide we can produce 0.351 moles of hydroxide. Let's calculate the mass:
0.351 mol . 99.4 g /1mol = 34.9 g
