Question

вычислить массу соли,получающуюся при нейтрализации азотной кислотой раствора,содержащего 4 г. гидроксида натрия

Answer 1

HNO3 + NaOH = NaNO3 + H2

1)n(NaOH) = m/Mr = 4/40 = 0,1 моль

2)n(NaNO3) = 0,1 моль

3)m(NaNO3) = 0,1*85= 8,5

Ответ: 8,5

Answer 2

Calculating the Mass of Salt Produced by Neutralizing Sodium Hydroxide with Nitric Acid

To calculate the mass of salt produced by neutralizing sodium hydroxide with nitric acid, we can use the following chemical equation:

2NaOH + HNO3 → NaNO3 + H2O

From the equation, we can see that 2 moles of sodium hydroxide react with 1 mole of nitric acid to produce 1 mole of sodium nitrate and 2 moles of water.

Given that the solution contains 4g of sodium hydroxide, we can calculate the mass of salt produced.

First, we need to find the molar mass of sodium hydroxide and nitric acid.

The molar mass of NaOH (sodium hydroxide) is 40 g/mol (sodium: 23g/mol, oxygen: 16g/mol, hydrogen: 1g/mol).

The molar mass of HNO3 (nitric acid) is 63 g/mol (nitrogen: 14g/mol, oxygen: 48g/mol, hydrogen: 1g/mol).

Now, we can calculate the amount of sodium hydroxide in moles:

moles of NaOH = mass of NaOH / molar mass of NaOH moles of NaOH = 4g / 40 g/mol moles of NaOH = 0.1 mol

According to the balanced chemical equation, 2 moles of NaOH react with 1 mole of HNO3 to produce 1 mole of NaNO3.

So, the moles of HNO3 required for the reaction will be half of the moles of NaOH:

moles of HNO3 = 0.1 mol / 2 moles of HNO3 = 0.05 mol

Now, we can calculate the mass of NaNO3 (sodium nitrate) produced:

mass of NaNO3 = moles of NaNO3 * molar mass of NaNO3 mass of NaNO3 = 0.05 mol * 85 g/mol mass of NaNO3 = 4.25g

Therefore, the mass of salt (sodium nitrate) produced by neutralizing 4g of sodium hydroxide with nitric acid is 4.25g.