How many grams of NaCl would be required
to react with 346 mL of 0.607 M AgNO3
solution?

Mass = 12.27 g

Explanation:

Given data:

Mass of NaCl are required = ?

Volume of AgNO₃ = 346 mL (346 mL× 1L/ 1000 mL = 0.346 L)

Molarity of AgNO₃ = 0.607 M

Solution:

Chemical equation:

NaCl  + AgNO₃     →    NaNO₃ + AgCl

Number of moles of AgNO₃:

Molarity = number of moles / volume in L

0.607 M = number of moles / 0.346 L

number of moles =  0.607 M × 0.346 L

number of moles =  0.21 mol

now we will compare the moles of NaCl and AgNO₃.

                       AgNO₃         :          NaCl        

                           1                :            1

                       0.21              :            0.21

Mass of NaCl:

Mass = number of moles × molar mass

Mass = 0.21 mol × 58.44 g/mol

Mass = 12.27 g