15. 120 g of sulfuric acid are added to 230.0 g of barium peroxide. Which reactant is in excess and by how much? How much barium sulfate is precipitated in this reaction????
HELP please
Here is an example of a limiting reagent problem. Just follow the steps.
To determine which reactant is in excess and by how much, we need to compare the moles of sulfuric acid and barium peroxide.
First, we need to calculate the number of moles for sulfuric acid and barium peroxide. We can do this by using the formula:
Number of moles = Mass (g) / Molar mass
The molar mass of sulfuric acid (H2SO4) is calculated as follows:
2(1.01 g/mol for hydrogen) + 32.06 g/mol (for sulfur) + 4(16.00 g/mol for oxygen) = 98.09 g/mol
Number of moles of sulfuric acid = 120 g / 98.09 g/mol
Similarly, the molar mass of barium peroxide (BaO2) is calculated as follows:
1(137.33 g/mol for barium) + 2(16.00 g/mol for oxygen) = 169.33 g/mol
Number of moles of barium peroxide = 230.0 g / 169.33 g/mol
Now that we have calculated the number of moles for each reactant, we can compare them.
If the moles of sulfuric acid are greater than the moles of barium peroxide, then sulfuric acid is in excess. If the moles of barium peroxide are greater, then barium peroxide is in excess.
To calculate the amount of barium sulfate precipitated, you need to consider the balanced chemical equation and stoichiometry of the reaction. Unfortunately, you have not provided the balanced equation for the reaction between sulfuric acid and barium peroxide. Can you please provide it?