amount of heat envolved when 500 ml of 0.1 M hcl is mixed with 200 ml of 0.2 NAOH
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Answer:
HCl+NaOH→NaCl+H2O
at t=0,
number of moles = 500×0.11000=200×0.21000
=0.05 =0.04
during neutralisation of 1 mole of NaOH by 1 mole of HCl, heat ecolved = 57. 3 kJ
to neutralised 0.04 moles of NaOH by 0.04 molw of NaOH, heat evolved
57.3×0.04
= 2.292 kJ
Explanation:
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