Question & Answer: What mass (in g) of AgCl is formed from the reaction of 75.0 mL of a 0.078 M AgC_2H_3O_2…..

47) What mass (in g) of AgCl is formed from the reaction of 75.0 mL of a 0.078 M AgC2H302 solution with 55.0 mL of 0.109 M MgCl2 solution? 2 AgC2H3O2(aq) + MgCl2(aq) → 2 AgCl(s) + Mg(C2H3O2)2(aq) A) 0.838 g C) 0.859 g B) 1.72 g D) 2.56 g E) 1.70 g

What mass (in g) of AgCl is formed from the reaction of 75.0 mL of a 0.078 M AgC_2H_3O_2 solution with 55.0 mL of 0.109 M MgCl_2 solution? 2 AgC_2H_3O_2(aq) + MgCl_2(aq) rightarrow 2 AgCl(s) + Mg(C_2H_3O_2)_2(aq) A) 0.838 g B) 1.72 g C) 0.859 g D) 2.56 g E) 1.70 g

Expert Answer

Answer

2AgC₂H₃O₂(aq) + MgCl₂(aq) $rightarrow$ 2AgCl(s) + Mg(C₂H₃O₂)₂(aq)

In the above chemical reaction equation,

2 moles of AgC₂H₃O₂ reacts with 1 mole of MgCl₂(aq) to produce 2 moles of AgCl and 1 mole of Mg(C₂H₃O₂)₂.

Now,

Molarity = Moles / Liter

Moles = Molarity x Liter

Moles of AgC₂H₃O₂ in 75 mL of 0.078 M = 0.078 M x 0.075 L

= 0.005850 moles

Moles of MgCl₂ in 55 mL of 0.109 M = 0.109 M x 0.055 L

= 0.005995 moles

Since, moles of AgC₂H₃O₂ is less and hence is the limiting reagent.

Again, 2 moles of AgC₂H₃O₂ produce 2 moles of AgCl.

Or, 1 moles of AgC₂H₃O₂ produce 1 moles of AgCl.

Or, 0.005850 moles of AgC₂H₃O₂ produce 0.005850 moles of AgCl.

Molar mass of AgCl = 143.32 g/mol

So, 1 mole of AgCl = 143.32 g

Or, 0.005850 moles of AgCl = (0.005850 x 143.32) g

= 0.838422 g

= 0.838 g

Answer is (A).

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