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Chapter 4: Problem 71

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

Short Answer

Expert verified
The mass of AgCl precipitate is \(0.215 \, g\).

Step by step solution

01

Write the Balanced Chemical Reaction

Firstly, you need to know the balanced chemical reaction between \(CaCl_2\) and \(AgNO_3\). It is given as: \(CaCl_2 + 2AgNO_3 → 2AgCl↓ + Ca(NO_3)_2\)
02

Calculate the Number of Moles

To calculate the number of moles, you can use the formula \(Molarity = Moles \, of \, Solute/Liters \, of \, Solution\). So for \(CaCl_2\), the number of moles is \(0.150 \, mol/L * 0.030 \, L = 0.0045 \, mol\). Meanwhile, for \(AgNO_3\), the number of moles is \(0.100 \, mol/L * 0.015 \, L = 0.0015 \, mol\).
03

Identify the Limiting Reactant

From the balanced chemical reaction, you can see that 1 mole of \(CaCl_2\) reacts with 2 moles of \(AgNO_3\). Hence, you have enough \(CaCl_2\) to react with all the \(AgNO_3\), but not enough \(AgNO_3\) to react with all the \(CaCl_2\). This means that \(AgNO_3\) is the limiting reactant.
04

Calculate the Moles of AgCl

The balanced equation tells us that 1 mole of \(AgNO_3\) will produce 1 mole of \(AgCl\). As \(AgNO_3\) is the limiting reactant with 0.0015 mol, you will have \(0.0015 \, mol\) of \(AgCl\) produced.
05

Convert the Moles of AgCl to Grams

The molar mass of \(AgCl\) is about \(143.32 \, g/mol\). To find the mass, you can use the formula \(Mass = Molar mass * Number of moles = 143.32 \, g/mol * 0.0015 \, mol = 0.215 \, g\).

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Most popular questions from this chapter

What is the advantage of writing net ionic equations?

What is the difference between a nonelectrolyte and an electrolyte? Between a weak electrolyte and a strong electrolyte?

Give oxidation numbers for the underlined atoms in the following molecules and ions: (a) \(\underline{\mathrm{Cs}_{2} \mathrm{O}},\) (b) \(\mathrm{Ca} \underline{\mathrm{I}}_{2}\), (c) \(\underline{\mathrm{Al}_{2} \mathrm{O}_{3}}\) (d) \(\mathrm{H}_{3} \mathrm{As} \mathrm{O}_{3},\) (e) \(\underline{\mathrm{Ti} \mathrm{O}_{2}},\) (f) \(\underline{\mathrm{Mo}} \mathrm{O}_{4}^{2-}\), (g) \(\underline{\mathrm{Pt} \mathrm{Cl}_{4}^{2-}}\) (h) \(\underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\) (i) \(\underline{\operatorname{Sn} F}_{2}\) (j) \(\underline{\mathrm{ClF}_{3},}\) (k) \(\underline{\mathrm{Sb}} \mathrm{F}_{6}^{-}\)

Based on oxidation number, explain why carbon monoxide (CO) is flammable but carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is not.

How many moles of \(\mathrm{MgCl}_{2}\) are present in \(60.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{MgCl}_{2}\) solution?
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