Silver(I) nitrate and sodium sulfate react to form silver(I) sulfate and sodium nitrate.

Suppose 43.5 mL of 0.040 M silver(I) nitrate solution and 26.58 mL of 0.10 M sodium sulfate solution are mixed. What is the maximum number of grams of silver(I) sulfate that can be made?

moles silver: .0435*.04

moles sulfate:.02658*.1

To make Ag2SO4, from the above, one is limiting. So of the two inputs above, which will be limiting?

silver

To find the maximum number of grams of silver(I) sulfate that can be made, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

Let's first calculate the number of moles of each reactant:

For silver(I) nitrate:
moles of AgNO3 = volume of AgNO3 solution (L) x concentration of AgNO3 (M)
moles of AgNO3 = (43.5 mL / 1000 mL/L) x 0.040 M
moles of AgNO3 = 0.00174 mol

For sodium sulfate:
moles of Na2SO4 = volume of Na2SO4 solution (L) x concentration of Na2SO4 (M)
moles of Na2SO4 = (26.58 mL / 1000 mL/L) x 0.10 M
moles of Na2SO4 = 0.00266 mol

Next, we need to determine the stoichiometric ratio between the reactants and the product. From the balanced chemical equation, we can see that the mole ratio between AgNO3 and Ag2SO4 is 1:1.

Since the mole ratio is 1:1 and the moles of AgNO3 is less than the moles of Na2SO4, the limiting reactant is AgNO3. This means that AgNO3 will be completely consumed, and the amount of Ag2SO4 that can be formed will be determined by the moles of AgNO3.

Now we can calculate the maximum number of moles of Ag2SO4 that can be formed:

moles of Ag2SO4 = moles of AgNO3 = 0.00174 mol

Finally, we can calculate the maximum number of grams of Ag2SO4:

mass of Ag2SO4 = moles of Ag2SO4 x molar mass of Ag2SO4

The molar mass of Ag2SO4 can be calculated using the atomic masses of silver (Ag) and sulfur (S):

molar mass of Ag2SO4 = (2 x atomic mass of Ag) + atomic mass of S + (4 x atomic mass of O)

molar mass of Ag2SO4 = (2 x 107.87 g/mol) + 32.06 g/mol + (4 x 16.00 g/mol)
molar mass of Ag2SO4 = 311.87 g/mol

mass of Ag2SO4 = 0.00174 mol x 311.87 g/mol
mass of Ag2SO4 = 0.541 g

Therefore, the maximum number of grams of silver(I) sulfate that can be made is 0.541 grams.

To find the maximum number of grams of silver(I) sulfate that can be made, we need to determine which reactant is limiting. The limiting reactant is the one that will be completely consumed and determines the maximum amount of product that can be formed.

First, let's calculate the number of moles of each reactant using the given concentrations and volumes:

For silver(I) nitrate:
Volume = 43.5 mL = 43.5 * 10^(-3) L
Concentration = 0.040 M

Number of moles of silver(I) nitrate = volume * concentration
= (43.5 * 10^(-3)) L * 0.040 M

For sodium sulfate:
Volume = 26.58 mL = 26.58 * 10^(-3) L
Concentration = 0.10 M

Number of moles of sodium sulfate = volume * concentration
= (26.58 * 10^(-3)) L * 0.10 M

Now, let's compare the number of moles of each reactant. The balanced equation tells us that the ratio of moles of silver(I) nitrate to moles of sodium sulfate is 1:1.

If the number of moles of silver(I) nitrate is equal to or greater than the number of moles of sodium sulfate, then silver(I) nitrate is the limiting reactant. Otherwise, sodium sulfate is the limiting reactant.

Now, we can calculate the number of moles of silver(I) sulfate that can be formed assuming silver(I) nitrate is the limiting reactant:

From the balanced equation, we know that 1 mole of silver(I) nitrate reacts to form 1 mole of silver(I) sulfate.

Therefore, the maximum number of moles of silver(I) sulfate is equal to the number of moles of silver(I) nitrate.

Finally, we can calculate the mass of silver(I) sulfate formed using the molar mass of silver(I) sulfate:

Molar mass of silver(I) sulfate = atomic mass of silver (Ag) + atomic mass of sulfur (S) + 4 * atomic mass of oxygen (O)

Using the molar mass, the number of moles, and the formula:

Mass of silver(I) sulfate = number of moles * molar mass

This will give us the maximum number of grams of silver(I) sulfate that can be made.