How many mL of0.4M AgNO3 solution must be added to 200mL of0.4 M NaI in order to for 11.7 grams of AgI precipitate?
To find the volume of the 0.4 M AgNO3 solution needed, we can use the concept of stoichiometry and the balanced chemical equation for the formation of AgI precipitate from AgNO3 and NaI.
The balanced chemical equation is:
AgNO3 + NaI → AgI + NaNO3
From the equation, we can see that 1 mole of AgNO3 reacts with 1 mole of NaI to form 1 mole of AgI. This means that the stoichiometric ratio of AgNO3 to AgI is 1:1.
Step 1: Calculate the moles of AgI needed
Given that 11.7 grams of AgI precipitate is formed, we need to convert this mass to moles. The molar mass of AgI is calculated as:
Ag (107.87 g/mol) + I (126.90 g/mol) = 234.77 g/mol
Moles = Mass / Molar mass = 11.7 g / 234.77 g/mol = 0.0498 mol (rounded to four decimal places)
Step 2: Calculate the volume of AgNO3 solution needed
Since the stoichiometric ratio of AgNO3 to AgI is 1:1, the moles of AgNO3 needed will be the same as the moles of AgI, which is 0.0498 mol.
Now, we can use the definition of molarity to calculate the volume:
Molarity = Moles / Volume (in L)
Rearranging the formula, we have Volume (in L) = Moles / Molarity
For AgNO3, we have:
Volume (in L) = 0.0498 mol / 0.4 mol/L = 0.1245 L (rounded to four decimal places)
Step 3: Convert volume to mL
Since we want the volume in mL, we can multiply the volume in L by 1000:
Volume (in mL) = 0.1245 L × 1000 mL/L = 124.5 mL
Therefore, approximately 124.5 mL of the 0.4 M AgNO3 solution must be added to 200 mL of the 0.4 M NaI solution in order for 11.7 grams of AgI precipitate to form.