An excess of AgNO3 reacts with 185.5 mL of an AlCl3 solution to give 0.325 g of AgCl. What is the concentration, in moles per liter, of the AlCl3 solution? Must show your work on scratch paper to receive credit. AlCl3(aq) + 3 AgNO3(aq) → 3 AgCl(s) + Al(NO3)3(aq)

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jufsanabriasa jufsanabriasa · Answer 1 · 2020-06-28T02:43:46+02:00

Answer:

4.07x10⁻³M AlCl₃.

Explanation:

Based on the reaction:

AlCl₃(aq) + 3 AgNO₃(aq) → 3 AgCl(s) + Al(NO₃)₃(aq)

That means 1 mole of AlCl₃ reacts with 3 moles of AgNO₃ to produce 3 moles of AgCl.

As 0.325g of AgCl are produced. Moles of AgCl are (Molar mass AgCl: 143.32g/mol):

0.325g AgCl ₓ ( 1 mol / 143.32g) = 2.27x10⁻³ moles of AgCl

As 3 moles of AgCl are produced from 1 mole of AlCl₃, moles of AlCl₃ that produce 2.27x10⁻³ moles of AgCl are:

2.27x10⁻³ moles of AgCl ₓ (1 mole AlCl₃ / 3 moles AgCl) =

7.56x10⁻⁴ moles AlCl₃

As volume of the AlCl₃ solution that reacts is 185.5mL = 0.1855L, molar concentration of the solution is:

7.56x10⁻⁴ moles AlCl₃ / 0.1855L =