the balanced equation for the formation of ammonia is
N₂ + 3H₂ ---> 2NH₃
molar ratio of N₂ to NH₃ is 1:2
mass of N₂ reacted is 8.0 g
therefore number of N₂ moles reacted is - 8.0 g / 28 g/mol = 0.286 mol
according to the molar ratio,
1 mol of N₂ will react to give 2 mol of NH₃, assuming nitrogen is the limiting reactant
therefore 0.286 mol of N₂ should give - 2 x 0.286 mol = 0.572 mol of NH₃
therefore mass of NH₃ formed is - 0.572 mol x 17 g/mol = 9.72 g
a mass of 9.72 mol of NH₃ is formed
Answer:
ZnSO4 + 2LiNO3 → Zn(NO3)2 + Li2SO4
Explanation:
There's many resources on web that can assist you with this concept:
https://en.intl.chemicalaid.com/tools/equationbalancer.php
https://www.webqc.org/balance.php
The answer is D transition state. In the energy profile, the transition state is the highest point. For a reaction, the activation energy is the minimal energy needed to trigger a reaction. The reactants are the start of the reaction and the products are the end of the reaction.
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
Answer:
We need 0.375 mol of CH3OH to prepare the solution
Explanation:
For the problem they give us the following data:
Solution concentration 0,75 M
Mass of Solvent is 0,5Kg
knowing that the density of water is 1g / mL, we find the volume of water:
Now, find moles of 
are needed using the molarity equation:
therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M
