What mass of NaCl is needed to precipitate all the silver ions from 20.0 mL of 0.100 M AgNO3 solution?
Follow this example. moles = M x L.
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To determine the mass of NaCl needed to precipitate all the silver ions from the AgNO3 solution, we first need to understand the reaction taking place between NaCl and AgNO3.
The balanced equation for the reaction between NaCl and AgNO3 is:
AgNO3 + NaCl → AgCl + NaNO3
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of NaCl to produce 1 mole of AgCl.
To calculate the mass of NaCl required, we need to consider the molar ratio between AgNO3 and NaCl.
Step 1: Calculate the moles of AgNO3
To find the moles of AgNO3, we use the formula:
moles = concentration × volume (in liters)
= 0.100 M × 0.0200 L
= 0.00200 moles
Step 2: Use the mole ratio from the balanced equation
From the balanced equation, we know that one mole of AgNO3 is equivalent to one mole of NaCl. Therefore, the moles of NaCl needed will also be 0.00200 moles.
Step 3: Calculate the mass of NaCl
To find the mass of NaCl, we use the formula:
mass = moles × molar mass
= 0.00200 moles × (22.99 g/mol + 35.45 g/mol)
= 0.00200 moles × 58.44 g/mol
= 0.11688 g
Therefore, to precipitate all the silver ions from 20.0 mL of 0.100 M AgNO3 solution, you would need approximately 0.11688 grams of NaCl.