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3 years ago

When solutions of barium chloride and lithium sulfate are mixed, the spectator ions in the resulting reaction are?

2 answers:

8 0

Barium chloride and lithium sulfate are mixed, barium sulfate and lithium chloride are formed. Barium sulfate is obtained as a white precipitate and lithium chloride as an aqueous solution.

Hope it helped!

ankoles [38]3 years ago

3 0

Answer : The resulting reaction will be,

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

$BaCl_2(aq)+Li_2SO_4(aq)\rightarrow BaSO_4(s)+2LiCl(aq)$

The ionic equation in separated aqueous solution will be,

$Ba^{2+}(aq)+2Cl^-(aq)+2Li^+(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2Li^+(aq)+2Cl^-(aq)$

In this equation, $Li^+\text{ and }Cl^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

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3 years ago

Heyy guys, so basically i need help with stoichiometric calculation I will give you 100 points just to answer all of these answe

jeka94

Answer:

3. The mass of ethanol required is approximately 0.522869 g

The mass of ethanoic acid required is approximately 0.68156 g

4. The mass of iron (III) oxide required is approximately 285.952.189.095 tonnes

5. The mass of silver nitrate required is approximately 14.53 grams

6. The mass of copper oxide that would be needed is approximately 31.86 grams

7. a. The mass of the precipitate, Zn(OH)₂ formed is approximately 49.712 grams

b. The mass of the precipitate, Al(OH)₃ formed is approximately 13 grams

c. The mass of the precipitate, Mg(OH)₂, formed is approximately 14.579925 grams

Explanation:

3. The 1 mole of ethanol and 1 mole of ethanoic acid combines to form 1 mole of ethyl ethanoate

The number of moles of ethyl ethanoate in 1 gram of ethyl ethanoate, n = 1 g/(88.11 g/mol) = 1/88.11 moles

∴ The number of moles of ethanol = 1/88.11 moles

The number of moles of ethanoic acid = 1/88.11 moles

The mass of ethanol = (46.07 g/mol) × 1/88.11 moles = 0.522869 g

The mass of ethanoic acid in the reaction = 60.052 g/mol × 1/88.11 moles ≈ 0.68156 g

4. 1 mole of iron(III) oxide reacts with 1 mole of CO₂ to produce 1 mole of iron

The number of moles in 100 tonnes of iron= 100000000/55.845 = 1790670.60614 moles

The mass of iron (III) oxide required = 159.69 × 1790670.60614 = 285952189.095 g ≈ 285.952.189.095 tonnes

5. The number of moles of NaCl in 5 grams of NaCl = 5 g/58.44 g/mol = 0.0855578371 moles

The mass of silver nitrate required, m = 169.87 g/mol × 0.0855578371 moles ≈ 14.53 grams

6. The number of moles of CuSO₄·5H₂O in 100 g of CuSO₄·5H₂O = 100 g/(249.69 g/mol) ≈ 0.4005 moles

The mass of copper oxide required, m = 79.545 g/mol × 0.4005 moles ≈ 31.86 grams

7. a. The number of moles of NaOH in the reaction = 20 g/(39.997 g/mol) ≈ 0.5 moles

2 moles of NaOH produces 1 mole of Zn(OH)₂

0.5 moles of NaOH will produce 0.5 mole of Zn(OH)₂

The mass of 0.5 mole of Zn(OH)₂ = 0.5 mole × 99.424 g/mol = 49.712 grams

The mass of the precipitate, Zn(OH)₂ formed = 49.712 grams

b. 6 moles of NaOH produces 2 moles Al(OH)₃

20 g, or 0.5 mole of NaOH will produce (1/6) mole of Al(OH)₃

The mass of the precipitate, Al(OH)₃ formed, m = 78 g/mol×(1/6) moles = 13 grams

c. 2 moles of NaOH produces 1 mole of Mg(OH)₂, therefore;

20 g or 0.5 moles of NaOH formed (1/4) mole of Mg(OH)₂

The mass of the precipitate, Mg(OH)₂, formed, m = 58.3197 g/mol × (1/4) moles = 14.579925 grams

3 0

2 years ago

Read 2 more answers

How many molecules are in 145.5 grams of Be(OH)2

Zanzabum

Answer:

2.04 x 10²⁴ molecules

Explanation:

Given parameters:

Mass of Be(OH)₂ = 145.5g

To calculate the number of molecules in this mass of Be(OH)₂ we follow the following steps:

>> Calculate the number of moles first using the formula below:

Number of moles = mass/molarmass

Since we have been given the mass, let us derive the molar mass of Be(OH)₂

Atomic mass of Be = 9g

O = 16g

H = 1g

Molar Mass = 9 + 2(16 + 1)

= 9 + 34

= 43g/mol

Number of moles = 145.5/43 = 3.38mol

>>> We know that a mole is the amount of substance that contains Avogadro’s number of particles. The particles can be atoms, molecules, particles etc. Therefore we use the expression below to determine the number of molecules in 3.38mol of Be(OH)₂:

Number of

molecules= number of moles x 6.02 x 10²³

Number of molecules= 3.38 x 6.02 x 10²³

= 20.37 x 10²³ molecules

= 2.04 x 10²⁴ molecules

3 0

2 years ago