Answer:
0.316 moles are produced.
Explanation:
We state the redox reaction:
4HNO₃ + Cu → Cu(NO₃)₂ + 2NO₂ + 2H₂O
We need to determine the limting reactant:
0.1 L . 14 M = 1.4 moles
10 g . 1mol/ 63.54g =0.158 mol
Cu is the limiting reactant. Let's see
4 moles of acid need 1 mol of Cu to react
1.4 moles of acid may react to (4 . 1) / 1.4 = 0.35 moles
We do not have enough Cu.
1 mol of Cu can produce 1 mol of NO₂
Then 0.158 moles will produce, 0.316 moles. (double of moles)
If we see stoichiometry, ratio is 1:2
Answer:
ΔH°f P4O10(s) = - 3115.795 KJ/mol
Explanation:
- P4O10(s) + 6H2O(l) ↔ 4H3PO4(aq)
- ΔH°rxn = ∑νiΔH°fi
∴ ΔH°rxn = - 327.2 KJ
∴ ΔH°f H2O(l) = - 285.84 KJ/mol
∴ ΔH°F H3PO4(aq) = - 1289.5088 KJ/mol
⇒ ΔH°rxn = (4)(- 1289.5088) - (6)(- 285.84) - ΔH°f P4O10(s) = - 327.2 KJ
⇒ ΔH°f P4O10(s) = - 5158.035 + 1715.04 + 327.2
⇒ ΔH°f P4O10(s) = - 3115.795 KJ/mol
According to valence bond theory sigma bonds is formed when two orbitals approach and overlap over each other while pie bonds is formed when two orbitals overlap side by side. in formation of HCl 1s orbital of hydrogen overlap on 3p orbitals of chlorine
Mr: 207.2
m=n×Mr= 6.53×207.2= 1353.02g
