Question

Расчитайте массу осадка образовавшейся при взаимодействии 5г NaOH с раствором CuSO4

Answer 1

      5                        x
2NaOH+CuSO₄=Cu(OH)₂↓+Na₂SO₄
   80                        98
x=5*98/80=6.125г

Answer 2

1)Составляем уравнение: 5г. x 2NaOH+CuSO4=Na2SO4+Cu(OH)2 80г/моль. 98г/моль 2) Находим M(молярную массу) M(NaOH)=23+16+1=40г/моль 2NaOH=40*2=80г/моль M(Cu(OH)2)=64+(16+1)*2=98г/моль 3) Составляем уравнение: 80*x=5*98 80x=490 x=6,125 Ответ:6,125 г.

Answer 3

Calculation of Precipitate Mass

To calculate the mass of the precipitate formed during the interaction of 5g of NaOH with a solution of CuSO4, we need to determine the stoichiometry of the reaction and use the balanced chemical equation.

The balanced chemical equation for the reaction between NaOH and CuSO4 is:

2NaOH + CuSO4 -> Na2SO4 + Cu(OH)2

From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of CuSO4 to produce 1 mole of Cu(OH)2.

To calculate the mass of the precipitate, we need to determine the number of moles of Cu(OH)2 formed. We can use the molar mass of Cu(OH)2 to convert moles to grams.

Let's calculate the mass of the precipitate step by step:

1. Calculate the number of moles of NaOH: - Given mass of NaOH = 5g - Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.99 g/mol - Number of moles of NaOH = mass / molar mass = 5g / 39.99 g/mol = 0.125 mol

2. Use the stoichiometry of the balanced equation to determine the number of moles of Cu(OH)2 formed: - From the balanced equation, we know that 2 moles of NaOH react with 1 mole of CuSO4 to produce 1 mole of Cu(OH)2. - Therefore, the number of moles of Cu(OH)2 formed is equal to half the number of moles of NaOH used. - Number of moles of Cu(OH)2 = 0.125 mol / 2 = 0.0625 mol

3. Calculate the mass of the precipitate (Cu(OH)2): - Molar mass of Cu(OH)2 = 63.55 g/mol + 1.01 g/mol + 16.00 g/mol = 80.56 g/mol - Mass of the precipitate = number of moles of Cu(OH)2 * molar mass = 0.0625 mol * 80.56 g/mol = 5.03 g

Therefore, the mass of the precipitate formed during the interaction of 5g of NaOH with a solution of CuSO4 is approximately 5.03 grams.