Question

Рассчитайте массу осадка, полученного при взаимодействии 24 г раствора MgSO4 c избытком раствора NaOH

Answer 1

MgSO4+NaOH=Mg(OH)2+Na2SO4 
Mg(OH)2 осадок 
M(MgSO4)=120г\моль
n(MgSO4)= m\M=24\120=0,2 моль
n(MgSO4):n(Mg(OH)2) = 1:1 => n(Mg(OH)2)= 0,2 моль
M(Mg(OH)2) = 58 г\моль
m=M x n= 58x0,2= 11,6г 

Answer 2

Calculating the Precipitate Mass

To calculate the mass of the precipitate obtained from the interaction of 24g of MgSO4 solution with an excess of NaOH solution, we can use the principles of stoichiometry and the concept of limiting reagents.

The balanced chemical equation for the reaction between MgSO4 and NaOH is:

MgSO4 + 2NaOH → Mg(OH)2 + Na2SO4

From the balanced equation, we can see that 1 mole of MgSO4 reacts with 2 moles of NaOH to produce 1 mole of Mg(OH)2.

Given that the molar mass of MgSO4 is 120.37 g/mol and the molar mass of Mg(OH)2 is 58.3197 g/mol, we can proceed with the calculation.

First, we need to determine the limiting reagent to find out how much Mg(OH)2 will be formed. Then, we can calculate the mass of the precipitate formed.

Let's proceed with the calculation.

Calculation Steps

1. Calculate the moles of MgSO4 using the given mass. 2. Use the mole ratio from the balanced equation to find the moles of Mg(OH)2 that can be formed. 3. Convert the moles of Mg(OH)2 to grams to find the mass of the precipitate.

Solution

1. Moles of MgSO4: - Moles = Mass / Molar mass - Moles = 24g / 120.37 g/mol - Moles ≈ 0.1995 mol

2. Moles of Mg(OH)2: - From the balanced equation, 1 mole of MgSO4 produces 1 mole of Mg(OH)2. - Therefore, the moles of Mg(OH)2 formed = 0.1995 mol

3. Mass of the Precipitate (Mg(OH)2): - Mass = Moles × Molar mass - Mass = 0.1995 mol × 58.3197 g/mol - Mass ≈ 11.63g

So, the mass of the precipitate obtained from the interaction of 24g of MgSO4 with an excess of NaOH solution is approximately 11.63 grams.