Question

Assume 0.38 L of a 2.8 M solution of magnesium chloride, MgCl₂, reacts with a 2.3 M solution of silver nitrate, AgNO₃, to produce silver chloride, AgCl, and magnesium nitrate, Mg(NO₃)₂. The problem requires that you determine the volume of silver nitrate, AgNO₃, needed for the reaction to occur. Which of the following options represents the correct ratio of the two solutions?

1) 2.3 M AgNO₃ over 2.8 M MgCl₂
2) 2.8 M MgCl₂ over 2.3 M AgNO₃
3) 2.3 mole AgNO₃ over 1 L AgNO₃
4) 1 L AgNO₃ over 2.3 mole AgNO₃

Answer 1

The correct ratio of the two solutions is 2.3 M AgNO₃ over 2.8 M MgCl₂. To determine the volume of silver nitrate needed for the reaction to occur, use the stoichiometry of the reaction and the given concentrations.

Step-by-step explanation:

The correct ratio of the two solutions is option 1) 2.3 M AgNO₃ over 2.8 M MgCl₂.

To determine the volume of silver nitrate needed for the reaction to occur, we need to use the stoichiometry of the reaction and the given concentrations. The balanced equation for the given reaction according to the question is:

MgCl₂ + 2AgNO₃ → 2AgCl + Mg(NO₃)₂

From the equation, we can see that 1 mole of MgCl₂ reacts with 2 moles of AgNO₃. Using the given concentration of 0.38 L of 2.8 M MgCl₂, we can calculate the moles of MgCl₂:

Moles of MgCl₂ = concentration * volume = 2.8 M * 0.38 L = 1.064 moles

Since the stoichiometry is 1:2 for MgCl₂ to AgNO₃, we need twice the moles of AgNO₃. Therefore, the moles of AgNO₃ needed = 2 * 1.064 moles = 2.128 moles.

Using the given concentration of the AgNO₃ solution, we can calculate the volume of AgNO₃ needed:

Volume of AgNO₃ = moles of AgNO₃ / concentration = 2.128 moles / 2.3 M = 0.926 L = 926 mL