What mass of NaCl is required to precipitate all the silver ions from 19.2 mL of 0.100 M AgNO3 solution?
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To find the mass of NaCl required to precipitate all the silver ions, we need to determine the stoichiometric ratio between silver ions (Ag+) and chloride ions (Cl-) in the reaction.
The balanced chemical equation for the reaction between silver nitrate (AgNO3) and sodium chloride (NaCl) is:
AgNO3 + NaCl -> AgCl + NaNO3
From the equation, we can see that 1 mole of silver nitrate reacts with 1 mole of sodium chloride to form 1 mole of silver chloride.
Now, let's calculate the number of moles of silver ions in the given volume of AgNO3 solution:
moles of Ag+ = concentration × volume
= 0.100 M × 0.0192 L
= 0.00192 moles
Since the stoichiometric ratio is 1:1 between Ag+ and Cl-, the number of moles of chloride ions required to precipitate all the silver ions is also 0.00192 moles.
Now, let's calculate the mass of NaCl required using the molar mass of NaCl:
mass of NaCl = moles × molar mass
= 0.00192 moles × (22.99 g/mol + 35.45 g/mol)
≈ 0.0882 g
Therefore, around 0.0882 grams of NaCl is required to precipitate all the silver ions from 19.2 mL of 0.100 M AgNO3 solution.
molessilver=.0192*.1=1.92E-3 moles
NaCl+AgNO3>AgCl+NaNO3
so it takes 1.92E-3 moles of salt, convert that to grams.