Ammonium sulfate reacts with a solution of sodium hydroxide to produce sodium sulfate, ammonia gas and water. Calculate the yield in grams of ammonia gas that would form; and the amount of excess reagent in grams that would be left in the flask, after 9.92 grams of ammonium sulfate reacts with 250mL of 0.5 M sodium hydroxide

This is a limiting reagent problem and you need to identify which reagent is the limiting reagent (LR). I know it is a LR problem because the amounts of BOTH reactants are given. I solve these by solving two simple stoichiometry problems (simple meaning they aren't limiting reagent problems). First use one reagent and find moles of the product. Second, use the other reagent and find moles of the product. The two answers probably will differ; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that smaller value is the limiting reagent. Here is an example problem of a simple stoichiometry problem.

http://www.jiskha.com/science/chemistry/stoichiometry.html
After you know the LR, use the coefficients between the LR and the other reagent to calculate the number of moles of that reagent used. subtract from the original amount, the convert to grams.

To calculate the yield of ammonia gas and the amount of excess reagent, we first need to write the balanced chemical equation for the reaction:

(NH4)2SO4 + 2NaOH → Na2SO4 + 2NH3 + 2H2O

From the equation, we can see that 1 mole of ammonium sulfate reacts with 2 moles of sodium hydroxide to produce 2 moles of ammonia gas.

Step 1: Calculate the moles of ammonium sulfate (NH4)2SO4.
Molar mass of (NH4)2SO4 = (2 x molar mass of NH4) + molar mass of SO4
= (2 x (14.01 + 1.01 x 4)) + (32.07 + 16.00 x 4)
= 132.14 g/mol

Moles of (NH4)2SO4 = mass / molar mass
= 9.92 g / 132.14 g/mol
= 0.075 moles

Step 2: Calculate the moles of sodium hydroxide (NaOH) from the given volume and concentration.
Moles of NaOH = volume (L) x concentration (mol/L)
= 250 mL x 0.5 mol/L
= 0.125 moles

Since the molar ratio between (NH4)2SO4 and NaOH is 1:2, we need twice the moles of NaOH.

Moles of NaOH = 0.125 moles x 2
= 0.250 moles

Step 3: Determine the limiting reagent.
The limiting reagent is the reactant that is completely consumed in the reaction, limiting the amount of product we can obtain. The reactant that yields fewer moles of product is the limiting reagent.

From the balanced equation, we can see that we need 1 mole of (NH4)2SO4 to react with 2 moles of NaOH. Therefore, the moles of (NH4)2SO4 should be twice the moles of NaOH for complete reaction. But in this case, the moles of (NH4)2SO4 (0.075 moles) are less than twice the moles of NaOH (0.250 moles).

Therefore, (NH4)2SO4 is the limiting reagent.

Step 4: Calculate the yield of ammonia gas (NH3).
From the balanced equation, we know that 1 mole of (NH4)2SO4 produces 2 moles of NH3.
So, moles of NH3 = moles of (NH4)2SO4 x 2
= 0.075 moles x 2
= 0.150 moles

To calculate the mass of NH3 in grams, we will use the molar mass of NH3 which is 14.01 g/mol.

Mass of NH3 = moles of NH3 x molar mass of NH3
= 0.150 moles x 14.01 g/mol
= 2.10 grams

Therefore, the yield of ammonia gas that would form is 2.10 grams.

Step 5: Calculate the excess reagent.
To determine the amount of excess reagent (sodium hydroxide) left in the flask, we need to find the moles of NaOH that were in excess.

Moles of NaOH in excess = Moles of NaOH initially - Moles of NaOH consumed
= 0.250 moles - (0.075 moles / 2)
= 0.250 moles - 0.0375 moles
= 0.2125 moles

Now, calculate the mass of excess NaOH using the molar mass which is 39.997 g/mol.

Mass of excess NaOH = Moles of NaOH in excess x Molar mass of NaOH
= 0.2125 moles x 39.997 g/mol
= 8.50 grams

Therefore, the amount of excess reagent (sodium hydroxide) left in the flask is 8.50 grams.