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Chapter 4: Problem 81

If \(30.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{CaCl}_{2}\) is added to \(15.0 \mathrm{~mL}\) of \(0.100 M \mathrm{AgNO}_{3},\) what is the mass in grams of \(\mathrm{AgCl}\) precipitate?

Short Answer

Expert verified
The mass of the \(\mathrm{AgCl}\) precipitate is \(0.215 \mathrm{~g}\).

Step by step solution

01

Write the balanced equation

The balanced equation of the reaction is: \[\mathrm{CaCl}_{2}(aq) + 2\mathrm{AgNO}_{3}(aq) \rightarrow 2\mathrm{AgCl}(s) + \mathrm{Ca(NO}_{3})_{2}(aq)\] Where the (s) stands for solid, indicating the precipitate, and the (aq) stands for aqueous, indicating the compounds that are dissolved in the solution.
02

Calculate moles of the reactants

The number of moles of \(\mathrm{CaCl}_{2}\) and \(\mathrm{AgNO}_{3}\) can be computed by multiplying the volume (in liters) by the molarity (moles per liter). So, \[\text{moles of } \mathrm{CaCl}_{2} = 0.030 \mathrm{~L} \times 0.150 \mathrm{~M} = 0.0045 \mathrm{~mol}\] \[\text{moles of } \mathrm{AgNO}_{3} = 0.015 \mathrm{~L} \times 0.100 \mathrm{~M} = 0.0015 \mathrm{~mol}\]
03

Identify the limiting reactant

According to the balanced equation, it takes one mole of \(\mathrm{CaCl}_{2}\) to react with two moles of \(\mathrm{AgNO}_{3}\), and form two moles of \(\mathrm{AgCl}\). Since you have more than enough \(\mathrm{CaCl}_{2}\) (0.0045 moles) for the \( \mathrm{AgNO}_{3}\) (0.0015 moles), the limiting reactant is \(\mathrm{AgNO}_{3}\). It will be completely consumed in the reaction.
04

Calculate the mass of the precipitate

The balanced equation tells you that two moles of \(\mathrm{AgCl}\) are formed for every two moles of \(\mathrm{AgNO}_{3}\) reacting (or one mole for one mole). So, it is expected that 0.0015 moles of \(\mathrm{AgCl}\) will be formed. The molar mass of \(\mathrm{AgCl}\) is 143.32 g/mol, hence the mass of the precipitate can be calculated as follows: \[\text{mass of } \mathrm{AgCl} = 0.0015 \mathrm{~mol} \times 143.32 \mathrm{~g/mol} = 0.215 g\]

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Most popular questions from this chapter

What volume of \(0.416 M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) should be added to \(255 \mathrm{~mL}\) of \(0.102 \mathrm{M} \mathrm{KNO}_{3}\) to produce a solution with a concentration of \(0.278 M \mathrm{NO}_{3}^{-}\) ions? Assume volumes are additive.

Write the equation for calculating molarity. Why is molarity a convenient concentration unit in chemistry?

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