- A mixture containing only BaO and CaO weighs 2.00 g. The oxides are converted to the corresponding mixed sulfates, which weigh 4.00 g. Calculate the percent Ba and Ca in the original mixture.
- A mixture containing only BaSO4 and CaSO4 contains one-half as much Ba2+ as Ca2+ by weight. What is the percentage of CaSO4 in the mixture?
The Correct Answer and Explanation is :
Problem 1:
Given:
- A mixture of BaO and CaO weighs 2.00 g.
- The mixture is converted to the corresponding mixed sulfates, BaSO₄ and CaSO₄, weighing 4.00 g.
Objective:
To calculate the percent of Ba and Ca in the original mixture of BaO and CaO.
Solution:
Let the masses of BaO and CaO in the mixture be represented as ( x ) and ( y ), respectively. So,
[
x + y = 2.00 \, \text{g}
]
The BaO reacts with sulfuric acid to form BaSO₄, and similarly, CaO reacts with sulfuric acid to form CaSO₄. The molar masses of BaO and CaO are 137.33 g/mol and 56.08 g/mol, respectively. The molar masses of BaSO₄ and CaSO₄ are 233.39 g/mol and 136.14 g/mol, respectively.
- The mass of BaSO₄ formed is related to the mass of BaO:
[
\text{Mass of BaSO₄} = \left( \frac{233.39}{137.33} \right) \times x
] - The mass of CaSO₄ formed is related to the mass of CaO:
[
\text{Mass of CaSO₄} = \left( \frac{136.14}{56.08} \right) \times y
]
The total mass of the sulfates is 4.00 g:
[
\left( \frac{233.39}{137.33} \right) \times x + \left( \frac{136.14}{56.08} \right) \times y = 4.00
]
Now, solve this system of equations:
- ( x + y = 2.00 )
- ( \left( \frac{233.39}{137.33} \right) \times x + \left( \frac{136.14}{56.08} \right) \times y = 4.00 )
From these, you can find ( x ) (mass of BaO) and ( y ) (mass of CaO). Once you have those, you can calculate the percentage of Ba and Ca in the original mixture.
Problem 2:
Given:
- A mixture contains BaSO₄ and CaSO₄.
- The mixture contains one-half as much Ba²⁺ as Ca²⁺ by weight.
Objective:
To find the percentage of CaSO₄ in the mixture.
Solution:
Let the mass of BaSO₄ be ( m_{\text{BaSO₄}} ) and the mass of CaSO₄ be ( m_{\text{CaSO₄}} ) in the mixture. The total mass is:
[
m_{\text{BaSO₄}} + m_{\text{CaSO₄}} = \text{Total mass of the mixture}
]
The amount of Ba²⁺ and Ca²⁺ in the mixture comes from the dissociation of BaSO₄ and CaSO₄.
- The molar mass of BaSO₄ is 233.39 g/mol, and Ba²⁺ makes up ( \frac{137.33}{233.39} ) of the molar mass.
- The molar mass of CaSO₄ is 136.14 g/mol, and Ca²⁺ makes up ( \frac{40.08}{136.14} ) of the molar mass.
The total mass of Ba²⁺ and Ca²⁺ is related to the masses of BaSO₄ and CaSO₄:
[
\frac{137.33}{233.39} \times m_{\text{BaSO₄}} = \frac{1}{2} \times \frac{40.08}{136.14} \times m_{\text{CaSO₄}}
]
From this, you can solve for the percentage of CaSO₄ in the mixture.
I’ll proceed with the calculations for both problems and also generate a relevant image.
Problem 1:
The mass of BaO in the mixture is approximately 1.175 g, and the mass of CaO is approximately 0.825 g.
- The percentage of Ba in the original mixture is 58.73%.
- The percentage of Ca in the original mixture is 41.27%.
Problem 2:
- The percentage of CaSO₄ in the mixture is approximately 79.99%.
Explanation:
For Problem 1, we used a system of equations to calculate the masses of BaO and CaO in the mixture. By using their respective molar mass ratios with BaSO₄ and CaSO₄, we were able to find the masses of each oxide in the mixture. The percentage of each component was then calculated based on the total mass of the mixture.
For Problem 2, we applied a mass balance for the Ba²⁺ and Ca²⁺ ions in the sulfates, considering that the mass of Ba²⁺ is half the mass of Ca²⁺. Using the molar masses of BaSO₄ and CaSO₄, we solved for the relative proportions of the sulfates in the mixture, yielding the percentage of CaSO₄.
I will now generate an image to visualize the process for both problems.
Here is the visual representation of the process for both problems, showing the conversion of BaO and CaO to BaSO₄ and CaSO₄, along with the calculated percentages. Let me know if you need further clarification!
