Question

Помогите с задачией заранее спасибоСколько граммов 5%-го раствора NaOH потребуется для нейтрализации 50 г 3%-ной азотной кислоты?

Answer 1

NaOH+HNO3=NaNO3+H2O
m(HNO3)=50*0.03=1.5г
n(HNO3)=1.5г:63г/моль=0,02моль
n(NaOH)=n(HNO3)=0.02моль
m(NaOH)=0.02*40=0.8г
m(всего раствора)=0,8:0,05=16г

Answer 2

Calculating the Amount of NaOH Needed to Neutralize Nitric Acid

To calculate the amount of 5% NaOH solution needed to neutralize 50g of 3% nitric acid, we can use the concept of equivalent weights and the equation:

NaOH + HNO3 → NaNO3 + H2O

First, we need to find the molar mass of NaOH and HNO3. Then we can use the equation to determine the amount of NaOH needed.

The molar mass of NaOH is 40 g/mol, and the molar mass of HNO3 is 63 g/mol.

Now, let's calculate the amount of NaOH needed.

Calculation

The molar concentration of a 5% NaOH solution is 5g/100ml. This means 100ml of the solution contains 5g of NaOH.

Given that the molar mass of NaOH is 40 g/mol, we can calculate the molarity of the 5% NaOH solution as follows:

Molarity = (mass / molar mass) / volume

Molarity = (5g / 40g/mol) / 0.1L

Molarity = 0.125 mol/L

Now, we can use the equation to find the amount of NaOH needed to neutralize 50g of 3% nitric acid.

The equation for the reaction is:

NaOH + HNO3 → NaNO3 + H2O

From the equation, we can see that 1 mole of NaOH reacts with 1 mole of HNO3.

The molar concentration of a 3% HNO3 solution is 3g/100ml. This means 100ml of the solution contains 3g of HNO3.

Given that the molar mass of HNO3 is 63 g/mol, we can calculate the molarity of the 3% HNO3 solution as follows:

Molarity = (3g / 63g/mol) / 0.1L

Molarity = 0.0476 mol/L

Now, we can calculate the amount of HNO3 in 50g of the 3% solution:

Amount of HNO3 = (50g / 63g/mol) = 0.7937 mol

Since the reaction is 1:1, we need the same amount of NaOH to neutralize the HNO3:

Amount of NaOH needed = 0.7937 mol

Finally, we can convert moles to grams:

Amount of NaOH needed = 0.7937 mol * 40g/mol = 31.748g

Answer

To neutralize 50g of 3% nitric acid, 31.748 grams of 5% NaOH solution are needed.