Question

До розчину масою 320 грам з масовою часткою натрій гідроксиду 20%, додали сульфатну кислоту. обчисліть масу солі , що утворилася .....

Answer 1

Ничего что на русском?

 

Реакция

2NaOH + H2SO4 = Na2SO4 + 2H2O

m(NaOH)=64 г

Mr(NaOH)=23+16+1=40 г\моль

M(NaOH)=1,6 моль

 

И до и после реакции - одинаковое количество атомов Na - 1,6 моль

До реакции он (Na) образует 1,6 моль вещества NaOH , после - 0,8 моль вещества Na2SO4 (это видно по реакции и формулам веществ).

 

Mr ( Na2SO4) = 46+32+64 = 142 г\моль

m(Na2SO4)= 142*0,8 = 113,6 г

 

Answer 2

Calculation of the Mass of Salt Formed

To calculate the mass of salt formed when 320 grams of a 20% sodium hydroxide solution is reacted with sulfuric acid, we need to determine the amount of sodium hydroxide present in the solution and the stoichiometry of the reaction.

First, let's calculate the mass of sodium hydroxide in the solution. Since the solution is 20% sodium hydroxide, we can calculate the mass of sodium hydroxide as follows:

Mass of sodium hydroxide = 20% of 320 grams = 0.20 * 320 grams = 64 grams.

Next, we need to determine the stoichiometry of the reaction between sodium hydroxide and sulfuric acid. The balanced chemical equation for the reaction is as follows:

2NaOH + H2SO4 -> Na2SO4 + 2H2O

From the balanced equation, we can see that 2 moles of sodium hydroxide react with 1 mole of sulfuric acid to form 1 mole of sodium sulfate and 2 moles of water.

Now, we can calculate the molar mass of sodium hydroxide (NaOH) and sulfuric acid (H2SO4):

- Molar mass of NaOH = 22.99 g/mol (sodium) + 16.00 g/mol (oxygen) + 1.01 g/mol (hydrogen) = 39.99 g/mol - Molar mass of H2SO4 = 1.01 g/mol (hydrogen) + 32.07 g/mol (sulfur) + 4 * 16.00 g/mol (oxygen) = 98.09 g/mol

Using the molar masses, we can calculate the number of moles of sodium hydroxide and sulfuric acid:

- Moles of NaOH = Mass of NaOH / Molar mass of NaOH = 64 g / 39.99 g/mol = 1.60 mol - Moles of H2SO4 = Moles of NaOH / 2 (according to the stoichiometry) = 1.60 mol / 2 = 0.80 mol

Since the stoichiometry of the reaction is 2:1 for NaOH to H2SO4, we can conclude that 0.80 moles of sodium hydroxide react with 0.80 moles of sulfuric acid.

Finally, we can calculate the mass of sodium sulfate (salt) formed using the molar mass of sodium sulfate (Na2SO4):

- Molar mass of Na2SO4 = 2 * (22.99 g/mol) (sodium) + 32.07 g/mol (sulfur) + 4 * 16.00 g/mol (oxygen) = 142.04 g/mol

The mass of sodium sulfate formed can be calculated as follows:

Mass of sodium sulfate = Moles of Na2SO4 * Molar mass of Na2SO4 = 0.80 mol * 142.04 g/mol = 113.63 grams.

Therefore, the mass of salt (sodium sulfate) formed is approximately 113.63 grams.

Please note that the above calculation assumes that the reaction goes to completion and there are no side reactions or losses during the process.