Question
Calculate the Mass of barium tetraoxosulophate (vi) that could be precipitated by a solution.containing 7.1g of disodium tetraoxosulphate (vi).
Answers
GPT-4o mini
To calculate the mass of barium tetraoxosulphate (VI) that could be precipitated by a solution containing 7.1 g of disodium tetraoxosulphate (VI), we first need to determine the relevant chemical equation for the reaction between barium ions and disodium tetraoxosulphate (VI).
The chemical formula for disodium tetraoxosulphate (VI) is \( \text{Na}_2\text{SO}_4 \). When this salt reacts with barium ions (\( \text{Ba}^{2+} \)), it forms barium tetraoxosulphate (VI), which has the formula \( \text{BaSO}_4 \).
The reaction can be written as follows:
\[
\text{Ba}^{2+} + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2\text{Na}^+
\]
### Step 1: Calculate the molar mass of disodium tetraoxosulphate (VI) \(\text{Na}_2\text{SO}_4\)
From the periodic table:
- Sodium (Na): approx. 23 g/mol
- Sulfur (S): approx. 32 g/mol
- Oxygen (O): approx. 16 g/mol
Molar mass of disodium tetraoxosulphate (VI):
\[
\text{Molar mass} = 2(\text{Na}) + \text{S} + 4(\text{O}) = 2(23) + 32 + 4(16) = 46 + 32 + 64 = 142 \text{ g/mol}
\]
### Step 2: Calculate the number of moles of disodium tetraoxosulphate (VI)
Using the molar mass, we can find the number of moles in 7.1 g of \( \text{Na}_2\text{SO}_4 \):
\[
\text{Moles of } \text{Na}_2\text{SO}_4 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{7.1 \text{ g}}{142 \text{ g/mol}} \approx 0.0500 \text{ mol}
\]
### Step 3: Determine the amount of barium tetraoxosulphate (VI) that can be precipitated
According to the equation, each mole of \( \text{Na}_2\text{SO}_4 \) yields one mole of \( \text{BaSO}_4 \). Thus, 0.0500 moles of \( \text{Na}_2\text{SO}_4 \) can produce 0.0500 moles of \( \text{BaSO}_4 \).
### Step 4: Calculate the molar mass of barium tetraoxosulphate (VI) \( \text{BaSO}_4 \)
From the periodic table:
- Barium (Ba): approx. 137 g/mol
- Sulfur (S): approx. 32 g/mol
- Oxygen (O): approx. 16 g/mol
Molar mass of barium tetraoxosulphate (VI):
\[
\text{Molar mass} = \text{Ba} + \text{S} + 4(\text{O}) = 137 + 32 + 4(16) = 137 + 32 + 64 = 233 \text{ g/mol}
\]
### Step 5: Calculate the mass of barium tetraoxosulphate (VI)
Now we can calculate the mass of \( \text{BaSO}_4 \) that can be formed from 0.0500 moles:
\[
\text{Mass of } \text{BaSO}_4 = \text{Moles} \times \text{Molar mass} = 0.0500 \text{ mol} \times 233 \text{ g/mol} \approx 11.65 \text{ g}
\]
### Conclusion
The mass of barium tetraoxosulphate (VI) that could be precipitated by a solution containing 7.1 g of disodium tetraoxosulphate (VI) is approximately **11.65 g**.
The chemical formula for disodium tetraoxosulphate (VI) is \( \text{Na}_2\text{SO}_4 \). When this salt reacts with barium ions (\( \text{Ba}^{2+} \)), it forms barium tetraoxosulphate (VI), which has the formula \( \text{BaSO}_4 \).
The reaction can be written as follows:
\[
\text{Ba}^{2+} + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2\text{Na}^+
\]
### Step 1: Calculate the molar mass of disodium tetraoxosulphate (VI) \(\text{Na}_2\text{SO}_4\)
From the periodic table:
- Sodium (Na): approx. 23 g/mol
- Sulfur (S): approx. 32 g/mol
- Oxygen (O): approx. 16 g/mol
Molar mass of disodium tetraoxosulphate (VI):
\[
\text{Molar mass} = 2(\text{Na}) + \text{S} + 4(\text{O}) = 2(23) + 32 + 4(16) = 46 + 32 + 64 = 142 \text{ g/mol}
\]
### Step 2: Calculate the number of moles of disodium tetraoxosulphate (VI)
Using the molar mass, we can find the number of moles in 7.1 g of \( \text{Na}_2\text{SO}_4 \):
\[
\text{Moles of } \text{Na}_2\text{SO}_4 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{7.1 \text{ g}}{142 \text{ g/mol}} \approx 0.0500 \text{ mol}
\]
### Step 3: Determine the amount of barium tetraoxosulphate (VI) that can be precipitated
According to the equation, each mole of \( \text{Na}_2\text{SO}_4 \) yields one mole of \( \text{BaSO}_4 \). Thus, 0.0500 moles of \( \text{Na}_2\text{SO}_4 \) can produce 0.0500 moles of \( \text{BaSO}_4 \).
### Step 4: Calculate the molar mass of barium tetraoxosulphate (VI) \( \text{BaSO}_4 \)
From the periodic table:
- Barium (Ba): approx. 137 g/mol
- Sulfur (S): approx. 32 g/mol
- Oxygen (O): approx. 16 g/mol
Molar mass of barium tetraoxosulphate (VI):
\[
\text{Molar mass} = \text{Ba} + \text{S} + 4(\text{O}) = 137 + 32 + 4(16) = 137 + 32 + 64 = 233 \text{ g/mol}
\]
### Step 5: Calculate the mass of barium tetraoxosulphate (VI)
Now we can calculate the mass of \( \text{BaSO}_4 \) that can be formed from 0.0500 moles:
\[
\text{Mass of } \text{BaSO}_4 = \text{Moles} \times \text{Molar mass} = 0.0500 \text{ mol} \times 233 \text{ g/mol} \approx 11.65 \text{ g}
\]
### Conclusion
The mass of barium tetraoxosulphate (VI) that could be precipitated by a solution containing 7.1 g of disodium tetraoxosulphate (VI) is approximately **11.65 g**.