If a solution containing 20.0 grams of nickle (11) chloride is mixed with an aqueous solution of containing 15.0 grams of silver nitrate,, determine the mount of silver chloride that theoretically will precipitate. How much nickle (11) nitrate is formed? Which reactant is limiting? Which reactant is in Excess? How much of the excess reactant remain after the reaction is complete?

the mass in the end is 35.0 grams

To determine the amount of silver chloride that will precipitate and the amount of nickel (II) nitrate formed, we need to follow these steps:

1. Write the balanced chemical equation for the reaction:
2AgNO3 + NiCl2 -> 2AgCl + Ni(NO3)2

2. Calculate the molar masses of each compound:
Molar mass of AgNO3 = 169.87 g/mol
Molar mass of NiCl2 = 129.59 g/mol
Molar mass of AgCl = 143.32 g/mol
Molar mass of Ni(NO3)2 = 164.85 g/mol

3. Convert the grams of each compound to moles:
Moles of NiCl2 = 20.0 g / 129.59 g/mol = 0.154 moles
Moles of AgNO3 = 15.0 g / 69.87 g/mol = 0.215 moles

4. Determine the stoichiometric ratio between the reactants and products based on the balanced equation:
From the equation, we see that 2 moles of AgNO3 react with 1 mole of NiCl2 to produce 2 moles of AgCl and 1 mole of Ni(NO3)2.

5. Calculate the amount of silver chloride that will precipitate:
Since 2 moles of AgNO3 react with 1 mole of NiCl2 to produce 2 moles of AgCl, we can use the mole ratio to find the moles of AgCl:
Moles of AgCl = (0.215 moles of AgNO3) * (2 moles of AgCl / 2 moles of AgNO3) = 0.215 moles

Now, convert the moles of AgCl to grams:
Mass of AgCl = 0.215 moles * 143.32 g/mol = 30.82 grams

Therefore, the amount of silver chloride that will theoretically precipitate is 30.82 grams.

6. Calculate the amount of nickel (II) nitrate formed:
Since 1 mole of NiCl2 reacts to produce 1 mole of Ni(NO3)2, we can use the mole ratio to find the moles of Ni(NO3)2:
Moles of Ni(NO3)2 = 0.154 moles

Now, convert the moles of Ni(NO3)2 to grams:
Mass of Ni(NO3)2 = 0.154 moles * 164.85 g/mol = 25.38 grams

Therefore, the amount of nickel (II) nitrate formed is 25.38 grams.

7. Determine the limiting reactant:
To determine the limiting reactant, we compare the ratios of the stoichiometric coefficients of the two reactants. The reactant that produces the smaller amount of the product is the limiting reactant.

From the balanced equation:
NiCl2:AgNO3 ratio = 1:2
Moles of NiCl2 = 0.154 moles
Moles of AgNO3 = 0.215 moles

Since the ratio of NiCl2:AgNO3 is 1:2, and we have less NiCl2 (0.154 moles) compared to AgNO3 (0.215 moles), NiCl2 is the limiting reactant.

8. Determine the excess reactant:
To find the excess reactant, calculate the moles of the excess reactant that were not used in the reaction.
Moles of excess AgNO3 = Moles of AgNO3 used - Moles of AgNO3 required
Moles of AgNO3 required = 2 * Moles of NiCl2 (based on the stoichiometric ratio)

Moles of AgNO3 required = 2 * 0.154 = 0.308 moles
Moles of excess AgNO3 = 0.215 - 0.308 = -0.093 moles

Since the moles of excess AgNO3 is negative, it means that there is no excess AgNO3 in this case.

9. Determine the amount of excess reactant remaining:
Since there is no excess AgNO3, the amount of excess reactant remaining is 0 grams.

To summarize:
- The theoretically calculated amount of silver chloride that will precipitate is 30.82 grams.
- The amount of nickel (II) nitrate formed is 25.38 grams.
- Nickel (II) chloride (NiCl2) is the limiting reactant.
- There is no excess reactant remaining for AgNO3, so the amount is 0 grams.