Answers: 2
and 
.
. So, the limiting reactant is, NaOH.
![q=n[(\Delta H_f^o\text{ of }CO_3^{2-})+(\Delta H_f^o\text{ of }H_2O)]-n[(\Delta H_f^o\text{ of }HCO_3^-)+(\Delta H_f^o\text{ of }OH^-)]](/tpl/images/0033/5458/8571b.png)
= -677 kJ/mole
= -286 kJ/mole
= -692 kJ/mole
= -230 kJ/mole![q=0.008mole[(-677kJ/mole)+(-286kJ/mole)]-0.008mole[(-692kJ/mole)+(-230kJ/mole)]](/tpl/images/0033/5458/15c05.png)
(conversion used : 1 kJ = 1000 J)
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40.0 ml of 0.200 m aqueous naoh is added to 200.0 ml of 0.100 m aqueous nahco3 in a flask maintained...
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