Silver chloride is formed by mixing silver nitrate and barium chloride solutions. What volume of 1.50 M barium chloride solution is needed to form 0.525 g of silver chloride

1.22 mL

Explanation:

Let's consider the following balanced reaction.

2 AgNO₃ + BaCl₂ ⇄ Ba(NO₃)₂ + 2 AgCl

The molar mass of silver chloride is 143.32 g/mol. The moles corresponding to 0.525 g are:

0.525 g × (1 mol/143.32 g) = 3.66 × 10⁻³ mol

The molar ratio of AgCl to BaCl₂ is 2:1. The moles  of BaCl₂ are 1/2 × 3.66 × 10⁻³ mol = 1.83 × 10⁻³ mol.

The volume of 1.50 M barium chloride containing 1.83 × 10⁻³ moles is:

1.83 × 10⁻³ mol × (1 L/1.50 mol) = 1.22 × 10⁻³ L = 1.22 mL

16

explanation:

the marks go by 4s