How many moles of sodium hydroxide are required to neutralize 0.20 mol of nitric acid?

Answer 1

Balanced Equation.

NaOH + HNO3 >> NaNO3 + H2O

Now, Molarity = moles of solute/liters of solution

0.800M HNO3 + mol/2.50L

mol of HNO3 = 2

2mol HNO3 (1mol NaOH/1molHNO3 )(39.998g NaOH/1mol NaOH )

= 79.996 grams

Answer 2

Molarity = moles of solute/Liters of solution

0.343 M NaOH = moles NaOH.2.5 Liters

= 0.8575 moles NaOH (39.998 grams/1 mole NaOH)

= 34 grams of NaOH

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Answer 3

First write the BALANCED reaction equation.

NaOH + HNO3 = NaNO3 + H2O

The molar ratios are 1:1::1:1

So be equivalence

it is 0.2:0.2:: 0.2 :0.2

So you require 0.2 moles NaOH

The reaction to completion will produce 0.2 moles (NaNO3) and 0.2 moles water.

Answer 4

NaOH + HNO3 --> H2O + NaNO3

(0.20 mol HNO3) x (1 mol NaOH) / (1 mol HNO3) = 0.20 mol NaOH

Answer 5

0,2 mol of NaOH is sufficient

Answer 6

0.20mol