For the following reaction, 77.9 grams of barium chloride are allowed to react with with 68.8 grams of potassium sulfate.

What is the maximum amount of barium sulfate that can be formed in grams?

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete in grams?

For limiting reagent problems I work two of the "simple" stoichiometry problems (simple defined as not limiting reagent). Use one reactant and calculate the moles of product, then the other reactant and calculate the mole of product. You will obtain two answers, of course, but the correct answer is ALWAYS the smaller value. Here is a worked example of a simple stoichiometry problem.

http://www.jiskha.com/science/chemistry/stoichiometry.html
Use the non-limiting reagent to determine how much of the limiting reagent is used and subtract from the original amount to determine the amount of LR not reacted.

To determine the maximum amount of barium sulfate that can be formed, we need to identify the limiting reagent first. The limiting reagent is the reactant that is completely consumed in a chemical reaction and dictates the amount of product that can be formed.

1. Convert the given masses of barium chloride and potassium sulfate to moles. To do this, divide the mass of each reactant by its molar mass. The molar mass of barium chloride, BaCl2, is 137.33 g/mol, and the molar mass of potassium sulfate, K2SO4, is 174.26 g/mol.

For barium chloride:
77.9 g / 137.33 g/mol = 0.568 mol

For potassium sulfate:
68.8 g / 174.26 g/mol = 0.395 mol

2. Write and balance the balanced chemical equation for the reaction, which is:

BaCl2 + K2SO4 -> BaSO4 + 2KCl

According to the balanced equation, one mole of barium chloride reacts with one mole of potassium sulfate to produce one mole of barium sulfate.

3. Compare the mole ratios of the reactants in the balanced equation with the moles calculated in step 1. In this case, both the reactants have a 1:1 mole ratio.

4. Since the mole ratio of the reactants is 1:1, the limiting reagent is the one present in less moles. In this case, potassium sulfate (K2SO4) is the limiting reagent because it is present in fewer moles (0.395 mol) compared to barium chloride (0.568 mol).

5. To determine the maximum amount of barium sulfate that can be formed, we need to find the amount of barium sulfate that could be produced if all the limiting reagent is reacted. Since the mole ratio between barium sulfate and potassium sulfate is 1:1, the amount of barium sulfate formed is equal to the moles of limiting reagent.

0.395 mol * 233.43 g/mol (molar mass of BaSO4) = 92.65 g

Therefore, the maximum amount of barium sulfate that can be formed is 92.65 grams.

To determine the formula for the limiting reagent, we already established that potassium sulfate is the limiting reagent. Its formula is K2SO4.

Finally, to find the amount of excess reagent remaining after the reaction is complete, subtract the amount of limiting reagent used from the initial amount of excess reagent. In this case, we need to determine the amount of barium chloride (excess reagent) remaining.

Moles of excess reagent = Initial moles of excess reagent - Moles of limiting reagent used

0.568 mol (initial moles of BaCl2) - 0.395 mol (moles of K2SO4 used) = 0.173 mol

To find the mass of the excess reagent remaining, multiply the remaining moles of the excess reagent by its molar mass:

0.173 mol * 137.33 g/mol (molar mass of BaCl2) = 23.7 g

Therefore, 23.7 grams of barium chloride remains after the reaction is complete.