What mass of silver chloride can be produced from 1.27L of a 0.156M solution of silver nitrate?

m AgCl = 28.395 g

Explanation:

AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)

∴ [AgNO3] = 0.156 mol/L

∴ V = 1.27L

⇒ mol AgNO3 = 0.156 mol/L * 1.27 L = 0.19812 mol AgNO3

mass AgCL:

⇒ m AgCl = 0.19812 mol AgNO3 * mol AgCl/molAgNO3 * 143.32gAgCl/molAgCl

⇒ m AgCl = 28.395 g