What volume of 0.300M AgNO3 will react exactly with 40ml of NaCl producing AgCl and NaNo3?

Answer 1

AgNO3 MW = 169.87 g/mole
NaCl MW = 58.44 g/mole

reaction:

AgNO3(aq) + NaCl(aq) --> AgCl(s) + NaNO3(aq)

I'll assume you made a mistake and meant 40 g of NaCl

and your question is how many ml of 0.3M AgNO4 are needed to react with the 40g NaCl

First find the moles of NaCl

moles NaCl = 40 g * ( mole/58.44g ) = 0.6844 moles

then since its a 1:1 relationship

(0.300 moles AgNO3/1000ml) * xml = 0.6844 moles

solving ==> x ml = 2281.3 ml or 2.28 liters (0.3 M AgNO3)

This is a lot of AgNO3 so I will assume that in your question was missing some important data like 40 ml of 0.05 M NaCl or something like, then:

if that is the case do this to find the moles of NaCl

40 ml * (0.05 mole NaCl/1000ml) = 0.002 moles (0.05 M NaCl)

and proceed as above to find the volume of AgNO3

Answer 2

AgNO3 + NaCl <===> AgCl + NaNO3

Since everything is 1:1, it will take the same volume of AgNO3 to react with NaCl PROVIDED that they share the same concentration. If they don't, you have some calculating to do ^^

Answer 3

You'll need the same number of moles of AgNO3 as you have of NaCl. Since you don't give the concentration of your NaCl solution, I can't calculate the answer, but you should be able to. Just use the volume and molarity of your NaCl solution to calculate the moles of NaCl, and then divide that by the molarity of your AgNO3 solution to give you the volume you need.