How many milliliters of 0.165 M aluminum chloride are required to react completely with 35.0 mL of 0.210 M silver nitrate? The net ionic equation is:
Ag⁺(aq) + Cl^-(aq) $\rarr$ AgCl(s)
Hint: Consider that there are no coefficients necessary in the net ionic equation.

Question

How many milliliters of 0.165 M aluminum chloride are required to react completely with 35.0 mL of 0.210 M silver nitrate? The net ionic equation is:
Ag⁺(aq) + Cl^-(aq)

\rarr

AgCl(s)
Hint: Consider that there are no coefficients necessary in the net ionic equation.

Quizplus · Accepted Answer

The reaction involves a 1:1 mole ratio between Ag⁺ and Cl⁻. Calculate moles of Ag⁺: 0.210 M * 0.035 L = 0.00735 moles. Since AlCl₃ provides 3 Cl⁻ ions per molecule, moles of AlCl₃ needed = 0.00735 moles / 3 = 0.00245 moles. Volume of AlCl₃ = 0.00245 moles / 0.165 M = 0.0148 L = 14.8 mL.

How Many Milliliters of 0 →\rarr→ AgCl(s) Hint: Consider That There Are No Coefficients Necessary in to React

How Many Milliliters of 0 $\rarr$ AgCl(s)
Hint: Consider That There Are No Coefficients Necessary in to React