Question

Сколько граммов осадка образуется при взаимодействии 980 г 5%-го раствора сульфата меди (II) с необходимым количеством гидроксида натрия?

Answer 1

m = 980 * 0.05 =  49 грамм чистого сульфата меди , тогда : 

 

        49                              х 

CuSO4 + 2NaOH = Cu(OH)2 + Na2SO4

     160                               98

 х = 30.01 граммов 

Answer 2

Calculating the Amount of Precipitate Formed

To calculate the amount of precipitate formed when 980 g of a 5% solution of copper (II) sulfate reacts with the necessary amount of sodium hydroxide, we can use the given information and the chemical equation for the reaction.

The balanced chemical equation for the reaction between copper (II) sulfate and sodium hydroxide is:

CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4

From the equation, we can see that 1 mole of copper (II) sulfate reacts with 2 moles of sodium hydroxide to produce 1 mole of copper (II) hydroxide and 1 mole of sodium sulfate.

Using the molar mass of copper (II) sulfate (CuSO4) and sodium hydroxide (NaOH), we can calculate the amount of precipitate formed.

Calculation

1. Calculate the number of moles of copper (II) sulfate: - Given mass of copper (II) sulfate = 980 g - Molar mass of CuSO4 = 63.55 g/mol (Cu) + 32.07 g/mol (S) + 4 * 16.00 g/mol (O) = 159.61 g/mol - Moles of CuSO4 = Given mass / Molar mass = 980 g / 159.61 g/mol = 6.14 moles

2. Calculate the amount of sodium hydroxide required: - According to the balanced chemical equation, 1 mole of CuSO4 reacts with 2 moles of NaOH. - Moles of NaOH required = 2 * Moles of CuSO4 = 2 * 6.14 moles = 12.28 moles

3. Calculate the amount of precipitate formed: - According to the balanced chemical equation, 1 mole of CuSO4 forms 1 mole of Cu(OH)2. - Moles of Cu(OH)2 formed = Moles of CuSO4 = 6.14 moles - Given that the molar mass of Cu(OH)2 is 97.56 g/mol, the mass of Cu(OH)2 formed = Moles of Cu(OH)2 * Molar mass = 6.14 moles * 97.56 g/mol = 598.94 g

The amount of precipitate formed when 980 g of a 5% solution of copper (II) sulfate reacts with the necessary amount of sodium hydroxide is approximately 598.94 grams.