BaCl2 + Na2SO4 ? BaSO4 + 2NaCl

BaCl2 + Na2SO4 ? BaSO4 + 2NaCl

a. How many grams of barium chloride will be required to react with 5.00 g of sodium sulfate?

b. How many grams of barium chloride are required for the precipitation of 5.00 g of barium sulfate?

c. How many grams of barium chloride are needed to produce 5.00 g of sodium chloride?

d. How many grams of sodium sulfate are necessary for the precipitation of 5.00 g of barium chloride?

e. How many grams of sodium sulfate have been added to barium chloride if 5.00 g of barium sulfate is precipitated?

f. How many pounds of sodium sulfate are equivalent to 5.00 lb of sodium chloride?

g. How many pounds of barium sulfate are precipitated by 5.00 lb of barium chloride?

h. How many pounds of barium sulfate are precipitated by 5.00 lb of sodium sulfate?

i. How many pounds of barium sulfate are equivalent to 5.00 lb of sodium chloride?

The correct answer and explanation is :

We are given the reaction:

[
\text{BaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{NaCl}
]

This reaction is a double displacement reaction that forms barium sulfate (BaSO₄) as a precipitate. Let’s go through each part of the question one by one.

a. How many grams of barium chloride will be required to react with 5.00 g of sodium sulfate?

First, calculate the molar masses of the compounds involved:

• Molar mass of ( \text{Na}_2\text{SO}_4 ) = ( 2 \times 22.99 + 32.07 + 4 \times 16.00 = 142.05 \, \text{g/mol} )
• Molar mass of ( \text{BaCl}_2 ) = ( 137.33 + 2 \times 35.45 = 208.23 \, \text{g/mol} )

From the reaction, the stoichiometric ratio between sodium sulfate ((\text{Na}_2\text{SO}_4)) and barium chloride ((\text{BaCl}_2)) is 1:1. Thus, the number of moles of sodium sulfate will be equal to the number of moles of barium chloride needed for the reaction.

1. Calculate moles of sodium sulfate:

[
\text{moles of Na}_2\text{SO}_4 = \frac{5.00 \, \text{g}}{142.05 \, \text{g/mol}} = 0.0352 \, \text{mol}
]

2. Since the stoichiometry is 1:1, moles of barium chloride required = moles of sodium sulfate = 0.0352 mol.
3. Convert moles of barium chloride to grams:

[
\text{grams of BaCl}_2 = 0.0352 \, \text{mol} \times 208.23 \, \text{g/mol} = 7.33 \, \text{g}
]

So, 7.33 grams of barium chloride are required to react with 5.00 grams of sodium sulfate.

b. How many grams of barium chloride are required for the precipitation of 5.00 g of barium sulfate?

Next, we use the stoichiometry of the reaction, where 1 mole of BaCl₂ produces 1 mole of BaSO₄.

1. Molar mass of BaSO₄ = ( 137.33 + 32.07 + 4 \times 16.00 = 233.39 \, \text{g/mol} ).
2. Calculate moles of barium sulfate:

[
\text{moles of BaSO}_4 = \frac{5.00 \, \text{g}}{233.39 \, \text{g/mol}} = 0.0214 \, \text{mol}
]

3. From the stoichiometry of the reaction, moles of barium chloride required = moles of barium sulfate = 0.0214 mol.
4. Convert moles of barium chloride to grams:

[
\text{grams of BaCl}_2 = 0.0214 \, \text{mol} \times 208.23 \, \text{g/mol} = 4.46 \, \text{g}
]

So, 4.46 grams of barium chloride are required to produce 5.00 grams of barium sulfate.

c. How many grams of barium chloride are needed to produce 5.00 g of sodium chloride?

From the reaction, 1 mole of barium chloride produces 2 moles of sodium chloride. Thus, we need to calculate the number of moles of sodium chloride (NaCl) and then determine the corresponding amount of barium chloride.

1. Molar mass of NaCl = ( 22.99 + 35.45 = 58.44 \, \text{g/mol} ).
2. Calculate moles of sodium chloride:

[
\text{moles of NaCl} = \frac{5.00 \, \text{g}}{58.44 \, \text{g/mol}} = 0.0856 \, \text{mol}
]

3. From the stoichiometry, ( 1 \, \text{mol} \, \text{BaCl}_2 ) produces 2 moles of NaCl, so moles of barium chloride required = ( \frac{0.0856}{2} = 0.0428 \, \text{mol} ).
4. Convert moles of barium chloride to grams:

[
\text{grams of BaCl}_2 = 0.0428 \, \text{mol} \times 208.23 \, \text{g/mol} = 8.92 \, \text{g}
]

So, 8.92 grams of barium chloride are needed to produce 5.00 grams of sodium chloride.

d. How many grams of sodium sulfate are necessary for the precipitation of 5.00 g of barium chloride?

Using stoichiometry, 1 mole of sodium sulfate reacts with 1 mole of barium chloride. Therefore, the number of moles of barium chloride will be equal to the number of moles of sodium sulfate.

1. Molar mass of BaCl₂ = 208.23 g/mol.
2. Calculate moles of barium chloride:

[
\text{moles of BaCl}_2 = \frac{5.00 \, \text{g}}{208.23 \, \text{g/mol}} = 0.0240 \, \text{mol}
]

3. From the stoichiometry, moles of sodium sulfate required = moles of barium chloride = 0.0240 mol.
4. Molar mass of sodium sulfate = 142.05 g/mol.
5. Convert moles of sodium sulfate to grams:

[
\text{grams of Na}_2\text{SO}_4 = 0.0240 \, \text{mol} \times 142.05 \, \text{g/mol} = 3.41 \, \text{g}
]

So, 3.41 grams of sodium sulfate are necessary for the precipitation of 5.00 grams of barium chloride.

e. How many grams of sodium sulfate have been added to barium chloride if 5.00 g of barium sulfate is precipitated?

The stoichiometry of the reaction shows that 1 mole of sodium sulfate reacts with 1 mole of barium chloride to produce 1 mole of barium sulfate.

1. Molar mass of BaSO₄ = 233.39 g/mol.
2. Calculate moles of barium sulfate:

[
\text{moles of BaSO}_4 = \frac{5.00 \, \text{g}}{233.39 \, \text{g/mol}} = 0.0214 \, \text{mol}
]

3. From the stoichiometry, moles of sodium sulfate required = moles of barium sulfate = 0.0214 mol.
4. Molar mass of sodium sulfate = 142.05 g/mol.
5. Convert moles of sodium sulfate to grams:

[
\text{grams of Na}_2\text{SO}_4 = 0.0214 \, \text{mol} \times 142.05 \, \text{g/mol} = 3.04 \, \text{g}
]

So, 3.04 grams of sodium sulfate have been added to barium chloride if 5.00 grams of barium sulfate is precipitated.

f. How many pounds of sodium sulfate are equivalent to 5.00 lb of sodium chloride?

1. We know that the molar ratio between sodium sulfate and sodium chloride is 1:2. First, convert pounds of sodium chloride to grams:

[
5.00 \, \text{lb} = 5.00 \times 453.592 = 2267.96 \, \text{g}
]

2. Calculate moles of sodium chloride:

[
\text{moles of NaCl} = \frac{2267.96 \, \text{g}}{58.44 \, \text{g/mol}} = 38.85 \, \text{mol}
]

3. From the stoichiometry, moles of sodium sulfate required = ( \frac{38.85}{2} = 19.43 \, \text{mol} ).
4. Convert moles of sodium sulfate to grams:

[
\text{grams of Na}_2\text{SO}_4 = 19.43 \, \text{mol} \times 142.05 \, \text{g/mol} = 2752.85 \, \text{g}
]

5. Convert grams to pounds:

[
2752.85 \, \text{g} = \frac{2752.85}{453.592} = 6.06 \, \text{lb}
]

So, 6.06 pounds of sodium sulfate are equivalent to 5.00 lb of sodium chloride.

g. How many pounds of barium sulfate are precipitated by 5.00 lb of barium chloride?

1. Convert pounds of barium chloride to grams:

[
5.00 \, \text{lb} = 5.00 \times 453.592 = 2267.96 \, \text{g}
]

2. Calculate moles of barium chloride:

[
\text{moles of BaCl}_2 = \frac{2267.96 \, \text{g}}{208.23 \, \text{g/mol}} = 10.9 \, \text{mol}
]

3. From the stoichiometry, moles of barium sulfate produced = moles of barium chloride = 10.9 mol.
4. Molar mass of barium sulfate = 233.39 g/mol.
5. Convert moles of barium sulfate to grams:

[
\text{grams of BaSO}_4 = 10.9 \, \text{mol} \times 233.39 \, \text{g/mol} = 2543.9 \, \text{g}
]

6. Convert grams to pounds:

[
2543.9 \, \text{g} = \frac{2543.9}{453.592} = 5.61 \, \text{lb}
]

So, 5.61 pounds of barium sulfate are precipitated by 5.00 lb of barium chloride.

h. How many pounds of barium sulfate are precipitated by 5.00 lb of sodium sulfate?

1. Convert pounds of sodium sulfate to grams:

[
5.00 \, \text{lb} = 5.00 \times 453.592 = 2267.96 \, \text{g}
]

2. Calculate moles of sodium sulfate:

[
\text{moles of Na}_2\text{SO}_4 = \frac{2267.96 \, \text{g}}{142.05 \, \text{g/mol}} = 15.96 \, \text{mol}
]

3. From the stoichiometry, moles of barium sulfate produced = moles of sodium sulfate = 15.96 mol.
4. Molar mass of barium sulfate = 233.39 g/mol.
5. Convert moles of barium sulfate to grams:

[
\text{grams of BaSO}_4 = 15.96 \, \text{mol} \times 233.39 \, \text{g/mol} = 3726.4 \, \text{g}
]

6. Convert grams to pounds:

[
3726.4 \, \text{g} = \frac{3726.4}{453.592} = 8.21 \, \text{lb}
]

So, 8.21 pounds of barium sulfate are precipitated by 5.00 lb of sodium sulfate.

i. How many pounds of barium sulfate are equivalent to 5.00 lb of sodium chloride?

Using the stoichiometry, 2 moles of NaCl produce 1 mole of BaSO₄. Therefore, 5.00 lb of NaCl (2267.96 g) will produce half the number of moles of barium sulfate:

[
\text{moles of BaSO}_4 = \frac{38.85}{2} = 19.43 \, \text{mol}
]

Convert moles of BaSO₄ to grams and then to pounds:

[
\text{grams of BaSO}_4 = 19.43 \, \text{mol} \times 233.39 \, \text{g/mol} = 4531.1 \, \text{g}
]

[
4531.1 \, \text{g} = \frac{4531.1}{453.592} = 10.00 \, \text{lb}
]

So, 10.00 pounds of barium sulfate are equivalent to 5.00 lb of sodium chloride.