2 moles of aqueous Barium Nitrate is combined with 1 mole of aqueous Sodium Sulfate. Assuming 100% yield, what mass of Barium Sulfate precipitate would be produced. 1. 175.043 grams

2. 233.391 grams

3. 466.781 grams

4. 116.953 grams

Here is an example problem (worked). Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To determine the mass of Barium Sulfate precipitate produced in this reaction, we need to use the stoichiometry of the balanced chemical equation.

The balanced chemical equation for this reaction is:
Ba(NO3)2 + Na2SO4 -> BaSO4 + 2NaNO3

From the balanced equation, we can see that the ratio of Barium Nitrate to Barium Sulfate is 1:1. This means that 2 moles of Barium Nitrate will produce 2 moles of Barium Sulfate.

To calculate the mass of Barium Sulfate precipitate, we need to know the molar mass of Barium Sulfate, which is calculated by adding the atomic masses of its constituents. The atomic mass of Barium (Ba) is 137.33 g/mol, the atomic mass of Sulfur (S) is 32.07 g/mol, and the atomic mass of Oxygen (O) is 16.00 g/mol. Therefore, the molar mass of BaSO4 is:
(137.33 g/mol) + (32.07 g/mol) + (4 * 16.00 g/mol) = 233.39 g/mol

Now, we can calculate the mass of Barium Sulfate precipitate:
2 moles BaSO4 * (233.39 g/mol) = 466.78 g

Since the question assumes a 100% yield, the correct answer is option 3, which is 466.781 grams.