Balanced equation: 2Na(s) + Cl₂(g) ---> 2NaCl(s)
when we have STP conditions, we can use this conversion: 1 mol = 22.4 L
first, we have to convert grams to molecules using the molar mass, and then use mole to mole ratio from the balanced equation.
molar mass of Na= 23.0 g/mol
ratio: 2 mol Na= 1 mol Cl₂ (based on coefficients of balanced equation)
calculations:
Answer: 44g
Explanation: The formular for finding Moles is ;
Moles = Mass / Molar Mass or Formular Mass.
Base on this question; Moles = 10, Mass = 440g, and Formular Mass = ?
Making 'Formular Mass', subject of the formular; we thus have;
Formular mass = Mass / Moles = 440/ 10 = 44g
Elements of iodine are fluorine, chlorine, bromine, and astatine.
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 .
