Question

Две задачки по химии, помогите(( 1) К раствору хлорида магния массой 95г и массовой долей растворенного вещества 10% прилили избыток раствора гидроксида натрия. Чему равна масса выпавшего осадка? 2) Массовая доля серной кислоты в растворе равна 25%. Какая масса этого раствора потребуется для нейтрализации гидроксида бария в растворе массой 22,8 г и массовой долей растворенного вещества 1,5%?

Answer 1

1)MgCl2+2NaOH=2NaCl+Mg(OH)2
95×0,1=9,5г
9.5/95=0,1 моль MgCl и Mg(OH)2
масса осадка Mg(OH)2=0,1×58г/моль=5,8г
2)H2SO4+Ba(OH)2=BaSO4+2H2O
22.8×1,5/100=0,342г
0.342/171=0,002моль Ba(OH)2 и H2SO4
m(серн кисл)=0,002×98=0,196г
m(раств кисл)=0,196/0,25×100=4,9г

Answer 2

Task 1: Mass of Precipitate in Magnesium Chloride Solution

To find the mass of the precipitate formed when an excess of sodium hydroxide solution is added to a solution of magnesium chloride, we can use the following steps:

1. Calculate the moles of magnesium chloride. 2. Determine the limiting reagent. 3. Calculate the moles of the precipitate formed. 4. Convert moles to mass.

# Step 1: Calculate the Moles of Magnesium Chloride

The molar mass of magnesium chloride (MgCl2) is 95.211 g/mol. Given that the mass of the solution is 95 g and the mass fraction of the dissolved substance is 10%, we can calculate the mass of magnesium chloride in the solution.

Mass of magnesium chloride in the solution: \[95 \, \text{g} \times 0.10 = 9.5 \, \text{g}\]

Moles of magnesium chloride: \[9.5 \, \text{g} \div 95.211 \, \text{g/mol} = 0.0998 \, \text{mol}\]

# Step 2: Determine the Limiting Reagent

Next, we need to determine the limiting reagent by comparing the moles of magnesium chloride to the moles of sodium hydroxide.

# Step 3: Calculate the Moles of the Precipitate Formed

The balanced chemical equation for the reaction between magnesium chloride and sodium hydroxide is: \[MgCl_2 + 2NaOH \rightarrow Mg(OH)_2 + 2NaCl\]

From the balanced equation, we can see that 1 mole of magnesium chloride reacts with 2 moles of sodium hydroxide to form 1 mole of magnesium hydroxide. Therefore, the moles of magnesium hydroxide formed will be half the moles of magnesium chloride.

Moles of magnesium hydroxide formed: \[0.0998 \, \text{mol} \times \frac{1}{2} = 0.0499 \, \text{mol}\]

# Step 4: Convert Moles to Mass

The molar mass of magnesium hydroxide (Mg(OH)2) is 58.3197 g/mol. Using this molar mass, we can calculate the mass of the precipitate formed.

Mass of the precipitate (magnesium hydroxide): \[0.0499 \, \text{mol} \times 58.3197 \, \text{g/mol} = 2.92 \, \text{g}\]

Therefore, the mass of the precipitate (magnesium hydroxide) formed is 2.92 grams.

(#1)(#2)(#3)

Task 2: Neutralization of Barium Hydroxide with Sulfuric Acid

To find the mass of sulfuric acid solution needed to neutralize barium hydroxide in a given solution, we can use the following steps:

1. Calculate the moles of barium hydroxide. 2. Determine the moles of sulfuric acid needed for neutralization. 3. Convert moles to mass.

# Step 1: Calculate the Moles of Barium Hydroxide

The molar mass of barium hydroxide (Ba(OH)2) is 171.34 g/mol. Given that the mass of the solution is 22.8 g and the mass fraction of the dissolved substance is 1.5%, we can calculate the mass of barium hydroxide in the solution.

Mass of barium hydroxide in the solution: \[22.8 \, \text{g} \times 0.015 = 0.342 \, \text{g}\]

Moles of barium hydroxide: \[0.342 \, \text{g} \div 171.34 \, \text{g/mol} = 0.00199 \, \text{mol}\]

# Step 2: Determine the Moles of Sulfuric Acid Needed

The balanced chemical equation for the reaction between barium hydroxide and sulfuric acid is: \[Ba(OH)_2 + H_2SO_4 \rightarrow BaSO_4 + 2H_2O\]

From the balanced equation, we can see that 1 mole of barium hydroxide reacts with 1 mole of sulfuric acid to form 1 mole of barium sulfate. Therefore, the moles of sulfuric acid needed will be the same as the moles of barium hydroxide.

Moles of sulfuric acid needed: \[0.00199 \, \text{mol}\]

# Step 3: Convert Moles to Mass

The m