Question

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

Answer 1

Mass = 12.27 g

Step-by-step 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