What mass of Na2CrO4 is required to precipitate all of the silver ions from 71.0 mL of a 0.101 M solution of AgNO3?
AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2NaNO3
mols AgNO3 = M x L = ?
Using the coefficients in the balanced equation, convert mols AgNO3 to mols Na2CrO4.
Now convert mols Na2CrO4 to gram. g = mols x molar mass
Well, preparing to precipitate all those silver ions sounds like a real chemistry circus! So, let's do some math juggling, shall we?
First, we need to find the number of moles of silver ions in the given solution. To do that, we multiply the volume (71.0 mL) by the concentration (0.101 M), converting mL to liters:
71.0 mL × (1 L / 1000 mL) = 0.071 L
Moles of AgNO3 = (concentration in M) × (volume in L)
= 0.101 mol/L × 0.071 L
= 0.007171 mol
Na2CrO4 and AgNO3 have a 1:2 molar ratio according to the balanced equation for precipitation:
2Na2CrO4 + 3AgNO3 → Ag2CrO4 + 6NaNO3
So, we need twice as many moles of Na2CrO4 as there are moles of AgNO3. Let's calculate that:
Moles of Na2CrO4 required = 2 × moles of AgNO3
= 2 × 0.007171 mol
= 0.014342 mol
Now, we need to convert that into grams, using the molar mass of Na2CrO4:
Molar mass of Na2CrO4 = 22.99 g/mol (2 sodium atoms) + 52.00 g/mol (1 chromium atom) + 64.00 g/mol (4 oxygen atoms)
= 161 g/mol
Mass of Na2CrO4 required = moles × molar mass
= 0.014342 mol × 161 g/mol
= 2.305 g
So, you'll need approximately 2.305 grams of Na2CrO4 to precipitate all the silver ions in the given silver nitrate solution. Just remember, in the chemistry circus, precision is vital to avoid any funny reactions!
To determine the mass of Na2CrO4 required to precipitate all of the silver ions, we need to calculate the moles of silver ions present in the solution and then use the stoichiometry from the balanced chemical equation to determine the moles of Na2CrO4 needed.
The balanced chemical equation for the reaction between AgNO3 and Na2CrO4 is:
2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3
Based on the balanced equation, we can see that two moles of silver ions (Ag+) react with one mole of Na2CrO4 to form one mole of Ag2CrO4.
Step 1: Calculate the moles of silver ions present:
Moles of AgNO3 = concentration (mol/L) x volume (L)
Moles of AgNO3 = 0.101 mol/L x 0.0710 L
Moles of AgNO3 = 0.007171 mol
Step 2: Calculate the moles of Na2CrO4 needed:
According to the balanced equation, 2 moles of AgNO3 react with 1 mole of Na2CrO4. Therefore, the moles of Na2CrO4 needed will be half of the moles of AgNO3.
Moles of Na2CrO4 needed = 0.007171 mol / 2
Moles of Na2CrO4 needed = 0.0035855 mol
Step 3: Calculate the mass of Na2CrO4:
Molar mass of Na2CrO4 = 22.99 g/mol (Sodium) + 51.996 g/mol (Chromium) + (4 x 16.00 g/mol) (Oxygen)
Molar mass of Na2CrO4 = 161.97 g/mol
Mass of Na2CrO4 needed = Moles of Na2CrO4 x Molar mass of Na2CrO4
Mass of Na2CrO4 needed = 0.0035855 mol x 161.97 g/mol
Mass of Na2CrO4 needed = 0.5812 g
Therefore, approximately 0.5812 grams of Na2CrO4 is required to precipitate all of the silver ions from 71.0 mL of a 0.101 M solution of AgNO3.
To determine the mass of Na2CrO4 required to precipitate all of the silver ions from the AgNO3 solution, we can use the concept of stoichiometry and the balanced chemical equation for the precipitation reaction.
The balanced chemical equation for the reaction is:
2 AgNO3 + Na2CrO4 -> Ag2CrO4 + 2 NaNO3
From the balanced equation, we can see that 2 moles of AgNO3 reacts with 1 mole of Na2CrO4 to produce 1 mole of Ag2CrO4.
To solve this problem, follow these steps:
Step 1: Calculate the number of moles of AgNO3 in the given volume.
To do this, use the formula: moles = concentration x volume
moles of AgNO3 = 0.101 M x (71.0 mL / 1000 mL/1 L) = 0.007171 moles
Step 2: Convert moles of AgNO3 to moles of Na2CrO4.
From the balanced equation, we know that the mole ratio of AgNO3 to Na2CrO4 is 2:1.
moles of Na2CrO4 = (0.007171 moles AgNO3) x (1 mole Na2CrO4 / 2 moles AgNO3) = 0.003585 moles
Step 3: Calculate the mass of Na2CrO4 using its molar mass.
The molar mass of Na2CrO4 is 161.97 g/mol.
mass of Na2CrO4 = moles x molar mass = 0.003585 moles x 161.97 g/mol = 0.581 g
Therefore, approximately 0.581 grams of Na2CrO4 is required to precipitate all of the silver ions from the 71.0 mL of a 0.101 M solution of AgNO3.