Question

Вычисли, какая масса гидроксида марганца(II) выпадет в осадок при взаимодействии раствора сульфатамарганца(II) с раствором, содержащим 0,24моль гидроксида натрия.

Answer 1

Дано:
n(NaOH) = 0,24 моль
+MnSO₄
Найти:
Mn(OH)₂
Решение:
Mr(NaOH) = 40 г/моль (см. Таблицу Менделеева)
m(NaOH) = n(NaOH)*Mr(NaOH) = 0,24 моль * 40 г/моль = 9,6г
2NaOH + MnSO₄ = Mn(OH)₂ + Na₂SO₄
Mr(NaOH)по реакции = 40*2 = 80 г/моль
Mr(Mn(OH)₂) = 89 г/моль
m(Mn(OH)₂) = $\frac{9,6*89}{80} =$ 10,68г
Ответ: m(Mn(OH)₂) = 10,68г

Answer 2

Calculating the Mass of Manganese(II) Hydroxide Precipitate

To calculate the mass of manganese(II) hydroxide that will precipitate when a solution of manganese(II) sulfate reacts with a solution containing 0.24 moles of sodium hydroxide, we can use the balanced chemical equation for the reaction between manganese(II) sulfate and sodium hydroxide.

The balanced chemical equation for the reaction is: MnSO4 + 2NaOH → Mn(OH)2 + Na2SO4

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

Given that the solution contains 0.24 moles of sodium hydroxide, we can calculate the moles of manganese(II) hydroxide that will precipitate and then convert it to mass.

Calculation Steps

1. Calculate the moles of manganese(II) hydroxide precipitate: - Moles of Mn(OH)2 = 0.24 moles of NaOH / 2 - Moles of Mn(OH)2 = 0.12 moles

2. Convert moles of manganese(II) hydroxide to mass: - Mass = Moles × Molar mass of Mn(OH)2

Let's calculate the molar mass of manganese(II) hydroxide and then proceed with the calculation.