A 50 ml sample of 0.025M AgNo3 is mixed with 15 ml of 0.04 M CrCl3. How many grams of AgCl will precipitate? Please show steps

moles AgNO3 = M x L = ??

moles CrCl3 = M x L = ??

Using the coefficients in the balanced equation, convert moles AgNO3 to moles AgCl.
Same procedure, convert moles CrCl3 to moles AgCl.
The two answers likely will be different which means one of them is wrong; in limiting reagent problems the correct answer is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.

Then convert moles AgCl to grams. g = moles x molar mass.

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To determine the grams of AgCl that will precipitate, we need to calculate the limiting reagent and then use stoichiometry to convert the limiting reagent into the product.

First, let's calculate the moles of AgNO3 and CrCl3 using the given concentrations and volumes:

Moles of AgNO3 = concentration x volume
Moles of AgNO3 = 0.025 M x (50 ml / 1000 ml/1 L)
Moles of AgNO3 = 0.00125 mol

Moles of CrCl3 = concentration x volume
Moles of CrCl3 = 0.04 M x (15 ml / 1000 ml/1 L)
Moles of CrCl3 = 0.0006 mol

Next, we need to determine the limiting reagent. This can be done by comparing the molar ratio between the reactants:

AgNO3 + 3CrCl3 -> AgCl + 3Cr(NO3)3

From the balanced equation, we can see that 1 mole of AgNO3 reacts with 3 moles of CrCl3. The actual ratio we have is:

Moles of AgNO3 / Stoichiometric coefficient of AgNO3 = 0.00125 / 1 = 0.00125
Moles of CrCl3 / Stoichiometric coefficient of CrCl3 = 0.0006 / 3 = 0.0002

The ratio of AgNO3 to CrCl3 is approximately 6.25:1, which means that AgNO3 is in excess.

Therefore, CrCl3 is the limiting reagent.

Now, let's use the moles of CrCl3 to calculate the moles of AgCl that will precipitate:

From the balanced equation, we can see that 1 mole of CrCl3 produces 1 mole of AgCl.

Moles of AgCl = Moles of CrCl3
Moles of AgCl = 0.0002 mol

Finally, we can convert the moles of AgCl to grams using the molar mass of AgCl:

Molar mass of AgCl = 107.87 g/mol

Grams of AgCl = Moles of AgCl x Molar mass of AgCl
Grams of AgCl = 0.0002 mol x 107.87 g/mol
Grams of AgCl = 0.021574 g

Therefore, approximately 0.0216 grams of AgCl will precipitate.

To determine the grams of AgCl that will precipitate, we need to find the limiting reactant between AgNO3 and CrCl3. The limiting reactant is the one that is completely consumed, thus determining the maximum amount of product that can be formed.

First, let's calculate the number of moles of AgNO3 in the 50 ml solution. Remember that concentration is given in moles per liter (M):

Number of moles of AgNO3 = Volume (in liters) x Concentration
= 0.050 L x 0.025 mol/L
= 0.00125 mol

Next, calculate the number of moles of CrCl3 in the 15 ml solution:

Number of moles of CrCl3 = Volume (in liters) x Concentration
= 0.015 L x 0.04 mol/L
= 0.0006 mol

Now, we need to determine the stoichiometric ratio between AgNO3 and AgCl. Based on the balanced chemical equation, we know that 1 mole of AgNO3 produces 1 mole of AgCl.

Since the stoichiometric ratio is 1:1, the number of moles of AgCl formed will be equal to the number of moles of AgNO3.

Therefore, the number of moles of AgCl formed = 0.00125 mol

To find the mass of AgCl, we need to use the molar mass of AgCl. The molar mass of Ag is 107.87 g/mol, and the molar mass of Cl is 35.45 g/mol.

Molar mass of AgCl = Molar mass of Ag + Molar mass of Cl
= 107.87 g/mol + 35.45 g/mol
= 143.32 g/mol

Finally, calculate the mass of AgCl:

Mass of AgCl = Number of moles of AgCl x Molar mass of AgCl
= 0.00125 mol x 143.32 g/mol
= 0.1791 g

Therefore, approximately 0.1791 grams of AgCl will precipitate.