50.0mL of .10 silver nitrate is added to 50.0mL of .20M calcium chloride. A white precipitate forms. After the reaction is complete, calculate the amount of precipitate that formed.
Well, based on my extensive knowledge of chemistry and comedy, I'm afraid I don't have a hilarious answer for this one. We'll have to resort to some actual calculations.
First, let's write out the balanced chemical equation for the reaction:
AgNO3 + CaCl2 → AgCl + Ca(NO3)2
From the equation, we can see that 1 mole of AgNO3 reacts with 2 moles of CaCl2 to form 2 moles of AgCl.
Next, let's find the number of moles of silver nitrate (AgNO3) and calcium chloride (CaCl2) we have.
Moles of AgNO3 = (volume in liters) x (molarity)
= (50.0 mL/1000 mL/L) x (0.10 moles/L)
= 0.005 moles
Moles of CaCl2 = (volume in liters) x (molarity)
= (50.0 mL/1000 mL/L) x (0.20 moles/L)
= 0.010 moles
Since the reaction stoichiometry is 1:2 (AgNO3:CaCl2), the limiting reactant is AgNO3. This means that all of the AgNO3 will be consumed in the reaction, while some CaCl2 will be left over.
Now, let's calculate the number of moles of AgCl (the precipitate) that will form.
Moles of AgCl = 2 x (moles of AgNO3)
= 2 x 0.005 moles
= 0.010 moles
Finally, let's convert the moles of AgCl to grams using its molar mass:
Molar mass of AgCl = 107.87 g/mol
Mass of AgCl = (moles of AgCl) x (molar mass of AgCl)
= 0.010 moles x 107.87 g/mol
= 1.079 g
So, approximately 1.079 grams of precipitate, specifically AgCl, will form. Hope this helps, even if it wasn't as funny as you were expecting!
To calculate the amount of precipitate that formed, we need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed in the reaction and determines the amount of product that can be formed.
First, let's calculate the moles of silver nitrate and calcium chloride using their respective concentrations and volumes:
Moles of silver nitrate = concentration × volume
= 0.10 M × 0.050 L
= 0.005 mol
Moles of calcium chloride = concentration × volume
= 0.20 M × 0.050 L
= 0.010 mol
The balanced chemical equation for the reaction between silver nitrate and calcium chloride is:
AgNO3 + CaCl2 → AgCl + Ca(NO3)2
The stoichiometric ratio between silver nitrate and silver chloride is 1:1. Therefore, for every 1 mole of silver nitrate, 1 mole of silver chloride is produced.
Since the moles of silver nitrate and calcium chloride are both in a 1:1 ratio with the moles of silver chloride formed, it suggests that the limiting reagent is silver nitrate because it has the fewer number of moles.
Therefore, the amount of precipitate (silver chloride) that formed is equal to the moles of silver nitrate used, which is 0.005 mol. We can convert this to grams using the molar mass:
Mass of precipitate = moles × molar mass
= 0.005 mol × (AgCl molar mass)
= 0.005 mol × (107.87 g/mol)
= 0.539 g
So, the amount of precipitate that formed is 0.539 grams.
To calculate the amount of precipitate that formed, you need to know the balanced equation for the reaction between silver nitrate (AgNO3) and calcium chloride (CaCl2). From the given information, we can deduce that the reaction proceeds as follows:
AgNO3(aq) + CaCl2(aq) → AgCl(s) + Ca(NO3)2(aq)
According to the equation, 1 mole of silver nitrate reacts with 2 moles of calcium chloride to produce 1 mole of silver chloride (the precipitate) and 1 mole of calcium nitrate.
Now, we need to determine the limiting reagent to determine the amount of precipitate formed. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To find the limiting reagent, we need to compare the number of moles of each reactant to the stoichiometric ratios in the balanced equation.
The number of moles can be calculated using the formula:
moles = concentration (mol/L) × volume (L)
For silver nitrate:
moles of AgNO3 = 0.10 mol/L × 0.050 L = 0.005 mol
For calcium chloride:
moles of CaCl2 = 0.20 mol/L × 0.050 L = 0.010 mol
Now, we compare the moles of each reactant to the stoichiometric ratio. Since the ratio of AgNO3 to CaCl2 is 1:2, it means that for every 1 mole of AgNO3, we need 2 moles of CaCl2.
From the calculations, we can see that we have 0.005 moles of AgNO3 and 0.010 moles of CaCl2. As there is a 1:2 stoichiometric ratio between AgNO3 and CaCl2, it means that we have an excess of CaCl2. Therefore, AgNO3 is the limiting reagent.
Now that we know that AgNO3 is the limiting reagent, we can determine the number of moles of AgCl (the precipitate) that will form.
From the balanced equation, the stoichiometric ratio between AgNO3 and AgCl is 1:1. Thus, the amount of AgCl formed will be equal to the moles of AgNO3 used.
Therefore, the amount of AgCl precipitate that formed is 0.005 moles.
2AgNO3 + CaCl2 >>AgCl + Ca(NO3)2
you started with .05moles of silver nitrate and
.1 moles of calcium chloride
So the reaction is limited by the small amount of silver nitrate.
so the amount of silver chloride formed is 1/2*.05=.025 moles. Convert that to mass.
check my thinking.