What volume of 0.010M silver nitrate will exactly react with 20ml of 0.0080M socium chloride solution?
multiply 0.0080 by 20ml and divide it by 0.010 to get the volume of AgNO3 in ml
Be careful here. Setup is correct, but molarity = moles/volume in liters not milliliters.
20 x10^-3L *(0.0080)= moles of sodium chloride
solving for volume,
moles/molarity = volume of silver nitrate= moles of sodium chloride/molarity of silver nitrate
thank you
To determine the volume of 0.010M silver nitrate solution that will exactly react with 20ml of 0.0080M sodium chloride solution, we need to use the concept of stoichiometry.
First, let's write a balanced chemical equation for the reaction between silver nitrate (AgNO3) and sodium chloride (NaCl):
AgNO3 + NaCl → AgCl + NaNO3
From the equation, we can see that one mole of silver nitrate reacts with one mole of sodium chloride to produce one mole of silver chloride and one mole of sodium nitrate.
Now, let's calculate the number of moles of sodium chloride in the given 20ml of 0.0080M solution:
Molarity (M) = moles/volume (L)
0.0080M = moles/0.020L
moles = 0.0080M × 0.020L
= 0.00016 moles
Since the stoichiometry of the reaction is 1:1 between silver nitrate and sodium chloride, we know that 0.00016 moles of silver nitrate will react with 0.00016 moles of sodium chloride.
Now, let's calculate the volume of 0.010M silver nitrate:
Molarity (M) = moles/volume (L)
0.010M = 0.00016 moles/volume (L)
volume (L) = 0.00016 moles/0.010M
= 0.016 L
Finally, to convert the volume from liters to milliliters:
volume (ml) = 0.016 L × 1000 ml/L
= 16 ml
Therefore, 16ml of the 0.010M silver nitrate solution will exactly react with 20ml of the 0.0080M sodium chloride solution.