How many milliliters of 0.635M K2CrO4 are needed to precipitate all the silver in 410mL of 0.158M AgNO3 as AgCrO4 ?
2AgNO3 + K2CrO4 ==> Ag2CrO4 + 2KNO3
How many mols AgNO3 do you have? That's M x L = 0.158*0.410 = about 0.065 but you do it more accurately than that.
That requires 1/2 that K2CrO4 or about 0.032 (look at the coefficients in the balanced equation).
Then M K2CrO4 = mols K2CrO4/L K2CrO4. You have mols and M, solve for L and convert to mL
Well, let's find out! So, first things first, we know that AgCrO4 will precipitate when K2CrO4 reacts with AgNO3. To find out how many milliliters of K2CrO4 are needed, we can use mole-to-mole ratios.
The balanced equation for this reaction is:
2 AgNO3 + K2CrO4 → Ag2CrO4 + 2 KNO3
From the equation, we can see that 2 moles of AgNO3 react with 1 mole of K2CrO4 to form 1 mole of Ag2CrO4.
So, let's do some math! We'll start by calculating the number of moles of AgNO3 in 410mL of 0.158M AgNO3.
Moles of AgNO3 = volume (in liters) × concentration (in M)
Moles of AgNO3 = 0.410 L × 0.158 mol/L
Moles of AgNO3 = 0.06458 mol
Since the mole ratio between AgNO3 and K2CrO4 is 2:1, we need half as many moles of K2CrO4.
Moles of K2CrO4 = 0.06458 mol ÷ 2
Moles of K2CrO4 = 0.03229 mol
Now, we need to find out the volume of 0.635M K2CrO4 that contains 0.03229 moles.
Volume (in liters) = Moles ÷ concentration (in M)
Volume (in liters) = 0.03229 mol ÷ 0.635 mol/L
Volume (in liters) = 0.05089 L
Finally, we convert the volume to milliliters:
Volume (in mL) = 0.05089 L × 1000 mL/L
Volume (in mL) = 50.89 mL
So, approximately 50.89 milliliters of 0.635M K2CrO4 are needed to precipitate all the silver in 410mL of 0.158M AgNO3 as AgCrO4. Remember, these calculations are based on theoretical values, so it's always a good idea to double-check in the lab and maybe bring a clown with you for some extra humor! 🤡
To determine the number of milliliters of 0.635M K2CrO4 needed to precipitate all the silver in 410mL of 0.158M AgNO3 as AgCrO4, we need to use the stoichiometry of the balanced equation between K2CrO4 and AgNO3.
The balanced equation is as follows:
2 AgNO3 + K2CrO4 -> Ag2CrO4 + 2 KNO3
According to the equation, 2 moles of AgNO3 react with 1 mole of K2CrO4 to produce 1 mole of Ag2CrO4.
Step 1: Calculate the number of moles of AgNO3 in 410mL of 0.158M AgNO3.
moles of AgNO3 = volume (L) x concentration (mol/L)
moles of AgNO3 = (410 mL / 1000 mL/L) x 0.158 mol/L
Step 2: Convert moles of AgNO3 to moles of K2CrO4 using the stoichiometry of the balanced equation.
Since the stoichiometry between AgNO3 and K2CrO4 is 2:1, we divide the number of moles of AgNO3 by 2 to get the moles of K2CrO4 required.
moles of K2CrO4 = moles of AgNO3 / 2
Step 3: Convert moles of K2CrO4 to milliliters of 0.635M K2CrO4 using the concentration of K2CrO4.
volume (L) = moles / concentration (mol/L)
volume (mL) = volume (L) x 1000 mL/L
Now, let's perform the calculations:
Step 1:
moles of AgNO3 = (410 mL / 1000 mL/L) x 0.158 mol/L
moles of AgNO3 = 0.06458 mol
Step 2:
moles of K2CrO4 = 0.06458 mol AgNO3 / 2
moles of K2CrO4 = 0.03229 mol
Step 3:
volume (L) = 0.03229 mol / 0.635 mol/L
volume (mL) = 0.03229 L x 1000 mL/L
volume (mL) = 32.29 mL
Therefore, approximately 32.29 milliliters of 0.635M K2CrO4 are needed to precipitate all the silver in 410mL of 0.158M AgNO3 as AgCrO4.
To find the quantity of K2CrO4 needed to precipitate all the silver in the AgNO3 solution, we can use the concept of stoichiometry.
The balanced chemical equation for the reaction between K2CrO4 and AgNO3 is as follows:
2AgNO3 + K2CrO4 → Ag2CrO4 + 2KNO3
We can see from the equation that for every 2 moles of AgNO3, we need 1 mole of K2CrO4 to fully precipitate the silver as Ag2CrO4.
First, let's calculate the number of moles of AgNO3 in the 410 mL, 0.158 M solution:
Number of moles of AgNO3 = concentration (M) × volume (L)
= 0.158 mol/L × 0.410 L
= 0.06478 mol
Since we know that 2 moles of AgNO3 react with 1 mole of K2CrO4, we can use this ratio to find the number of moles of K2CrO4 needed:
Number of moles of K2CrO4 = 1/2 × number of moles of AgNO3
= 1/2 × 0.06478 mol
= 0.03239 mol
Now, let's convert the number of moles of K2CrO4 into milliliters using its molarity:
Molarity (M) = number of moles/volume (L)
Since we need to find the volume in milliliters, we rearrange the equation:
Volume (L) = number of moles/Molarity (M)
Volume (mL) = Volume (L) × 1000
Volume (mL) = 0.03239 mol / 0.635 mol/L × 1000
= 51.02 mL
Therefore, 51.02 mL of 0.635 M K2CrO4 is needed to precipitate all the silver in a 410 mL solution of 0.158 M AgNO3 as AgCrO4.