1) a) The reactions that occur are:
NaOH + HCl = NaCl + H2O
b) Na + + OH- + H + + Cl- = Na + + Cl- + H2O
c) OH- + H + = H2O
2) a) The heat of neutralization will be greater in Part A, than in Part B.
b) The molar heat of neutralization will be the same.
3) a) The reagent moles are calculated:
n NaOH = M * V = 2 M * 0.01 L = 0.02 mol
n HCl = 2 M * 0.01 L = 0.02 mol
b) Both are limit reagents. Well, both have the same amount of moles and the same stoichiometric coefficient.
c) 0.02 mol of NaCl are formed.
d) The resulting volume is 20 mL.
e) The molarity is calculated:
[NaCl] = n / V = 0.02 / 0.02 = 1 M
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