Question

When solutions of silver nitrate, AgNO3, and calcium iodide, CaI2, are mixed, a yellow precipitate of silver iodide is formed. Calculate the mass of silver iodide that is formed if 50.00 mL of 1.00 M AgNO3 is combined with 30.00 mL of 1.25 M CaI2.

Answer 1

Answer:

$\boxed{\text{11.7 g}}$

Explanation:

We are given the amounts of two reactant solutions, so this is a limiting reactant problem.

We know that we will need a balanced equation with masses, moles, and molar masses, so, let's assemble our information in one place, with molar masses above the formulas and the

M_r:                                                                              234.77

                      2AgNO₃          +           CaI₂               ⟶ 2AgI + Ca(NO₃)₂

Solution: (50.00 mL, 1.00 M)    (30.00 mL, 1.25 M)

Step 1.  Calculate the moles of each reactant  

Moles of AgNO₃ = 50.00 mL × (1.00 mmol/1 mL)  = 50.00 mmol

    Moles of CaI₂ = 30.00 mL × (1.25 mmol/1 mL)  = 37.50 mmol

 

Step 2. Identify the limiting reactant  

Calculate the moles of AgI we can obtain from each reactant.

From AgNO₃:  

The molar ratio of AgI:AgNO₃ is 2:2.

Moles of AgI  = 50.00 mmol AgNO₃ × (2 mmol AgI/2 mmol AgNO₃)

= 50.00 mmol AgI  

From CaI₂:

The molar ratio of AgI:CaI₂ is 2:1.  

Moles of CaI₂ = 37.50 mmol × (2 mmol AgI/1 mmol CaI₂) = 75.00 mmol AgI  

AgNO₃ is the limiting reactant because it gives the smaller amount of AgI.  

Step 3. Calculate the mass of AgI.

Mass = 50.00 mmol AgI × (234.77 mg AgI /1 mmol AgI)

= 11 700 mg AgI = 11.7 g AgI

The mass of silver iodide formed is $\boxed{\textbf{11.7 g}}$.