Question

what weight of Sodium Hydroxide can be neutralize 30 gram of nitric acid ​

Answer 1

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NaOH + HNO₃ → NaNO₃ + H₂O

One mole of NaOH (40 g) neutralises one mole of HNO₃ (63 g)

30 g of HNO₃ has: 30/63 = 0.48 moles of HNO₃

As one mole of HNO₃ requires one mole of NaOH,

Then, 0.48 moles of HNO₃ will require: 0.48 moles of NaOH

=> 0.48 x 40 = 19.2 g of NaOH

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