A 12 mL sample of 0.4 M silver nitrate solution is mixed with 8 mL of 0.8 M sodium chloride solution. Determine the identity of precipitate and the number of grams produced.
12 mL x 0.4M AgNO3 = 4.8 millimols.
8 mL x 0.8M NaCl = 6.4 mmols.
AgNO3 + NaCl => AgCl + NaNO3
This is a limiting reagent problem since amounts are given for BOTH reactants. You can tell, since the reactants and products are 1:1:1:1 that AgNO3 is the limiting reagent so you will produce 4.8 mmillimols (0.0048 mols) AgCl and convert that to grams. g = mols x molar mass.
Well, mixing silver nitrate and sodium chloride sounds like a recipe for a hilariously explosive experiment! But don't worry, no clowns were harmed in the making of this answer.
When you mix silver nitrate (AgNO3) and sodium chloride (NaCl), a precipitate called silver chloride (AgCl) is formed. It's a bit like a "Clown Party in a Flask"! But be careful, because silver chloride is no joke!
To determine the number of grams of precipitate produced, we need to use stoichiometry. The balanced equation for this reaction is:
AgNO3 + NaCl → AgCl + NaNO3
The molar ratio between AgNO3 and AgCl is 1:1, which means that for every 1 mole of AgNO3, we get 1 mole of AgCl.
First, let's calculate the number of moles of AgNO3. Given that the concentration of AgNO3 is 0.4 M and the volume is 12 mL, we can convert the volume to liters:
12 mL = 12/1000 = 0.012 L
Now we can calculate the moles of AgNO3:
moles of AgNO3 = concentration × volume
= 0.4 M × 0.012 L
= 0.0048 moles
Next, let's do the same calculation for the sodium chloride solution. The concentration is 0.8 M and the volume is 8 mL:
8 mL = 8/1000 = 0.008 L
moles of NaCl = concentration × volume
= 0.8 M × 0.008 L
= 0.0064 moles
Since the molar ratio between AgNO3 and AgCl is 1:1, the number of moles of AgCl produced will be the same as the number of moles of AgNO3 or NaCl (whichever is limiting). In this case, AgNO3 is limiting because it has fewer moles.
Therefore, the number of moles of AgCl produced is 0.0048 moles.
The molar mass of AgCl is approximately 143.32 g/mol. Let's use this to calculate the mass of AgCl:
mass of AgCl = number of moles × molar mass
= 0.0048 moles × 143.32 g/mol
≈ 0.688 g
So, the amount of silver chloride precipitate produced is approximately 0.688 grams.
I hope this "Clown Chemistry" lesson brightened your day a bit! Just remember, when the clowns get involved, anything can happen.
To determine the identity of the precipitate formed and the number of grams produced, we need to determine the reaction that occurs when silver nitrate (AgNO3) reacts with sodium chloride (NaCl):
AgNO3 + NaCl → AgCl + NaNO3
This reaction forms silver chloride (AgCl) as a precipitate.
To find the number of grams of AgCl produced, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.
1. Calculate the number of moles for each reactant:
Number of moles of AgNO3 = concentration × volume
Number of moles of AgNO3 = 0.4 M × 12 mL = 0.0048 moles
Number of moles of NaCl = concentration × volume
Number of moles of NaCl = 0.8 M × 8 mL = 0.0064 moles
2. The stoichiometric ratio of AgNO3 to AgCl is 1:1. Therefore, the same number of moles of AgNO3 will react with AgCl. Comparing the moles of AgNO3 (0.0048 moles) to the moles of NaCl (0.0064 moles), we see that AgNO3 is the limiting reactant.
3. Since AgNO3 is the limiting reactant, the number of moles of AgCl produced will be equal to the number of moles of AgNO3 reacted. Therefore, 0.0048 moles of AgCl will be produced.
4. Calculate the molar mass of AgCl:
Molar mass of AgCl = atomic mass of Ag + atomic mass of Cl
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
5. Calculate the mass of AgCl produced:
Mass of AgCl = number of moles × molar mass
Mass of AgCl = 0.0048 moles × 143.32 g/mol ≈ 0.69 grams
Therefore, the precipitate formed is silver chloride (AgCl), and approximately 0.69 grams of AgCl will be produced.
To determine the identity of the precipitate and the number of grams produced, we need to understand the concept of a chemical reaction called a precipitation reaction.
In this case, a precipitation reaction occurs when two aqueous salt solutions are mixed, resulting in the formation of an insoluble compound called a precipitate. We can determine the identity of the precipitate by using solubility rules.
First, let's write the balanced chemical equation for the reaction:
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
According to the equation, silver nitrate (AgNO3) reacts with sodium chloride (NaCl) to produce silver chloride (AgCl) and sodium nitrate (NaNO3).
Next, we need to determine if AgCl is insoluble in water. To do this, we can refer to solubility rules. In general, most chloride salts are soluble, except for salts of silver, lead, and mercury(I), which are insoluble. Therefore, AgCl is insoluble in water and, as a result, forms a precipitate.
Now, we can calculate the number of grams of AgCl precipitate produced. To do this, we need to determine the limiting reagent. The limiting reagent is the reactant that will be completely consumed in the reaction and will determine the maximum amount of product that can be formed.
To find the limiting reagent, we can compare the number of moles of each reactant:
For AgNO3:
0.4 M * 0.012 L = 0.0048 moles
For NaCl:
0.8 M * 0.008 L = 0.0064 moles
Since AgNO3 has fewer moles than NaCl, it is the limiting reagent.
Next, we need to determine the number of moles of AgCl formed. From the balanced equation, we know that the stoichiometric coefficient for AgCl is 1. Therefore, the number of moles of AgCl formed will be equal to the number of moles of AgNO3.
So, the number of moles of AgCl precipitate formed is 0.0048 moles.
Finally, we can calculate the mass of AgCl precipitate produced using its molar mass, which is 143.32 g/mol.
Mass of AgCl = Moles of AgCl * Molar mass of AgCl
Mass of AgCl = 0.0048 moles * 143.32 g/mol
Calculating this, we find that the mass of AgCl precipitate produced is approximately 0.688 grams.