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Sagot :
To help visualize what we have to do, let's start with what we want to calculate.
We want the mass of precipitate, which is the mass of AgCl.
[tex]m_{AgCl}[/tex]When we work with reaction, we need to follow the ratios of the reaction in number of moles, so if we want the mass of AgCl, we will need the number of moles of AgCl first and then use the molar mass og AgCl to calculate it:
[tex]\begin{gathered} M_{AgCl}=\frac{m_{AgCl}}{n_{AgCl}} \\ m_{AgCl}=M_{AgCl}\cdot n_{AgCl} \end{gathered}[/tex]So, since the reaction is in excess of NaCl, what will determine the amount produced of AgCl is the amount of AgNO₃ that reacted. So, we have to consider the sotichiometry. however, since both have the same coefficient, the amount that react of AgNO₃ will be the same amount of AgCl produced:
[tex]n_{AgCl}=n_{AgNO_{3}}[/tex]So, our equation becomes:
[tex]m_{AgCl}=M_{AgCl}\cdot n_{AgNO_3}[/tex]Since what we know is the volume that reacted and the concentration of the solution, we need to use the molarity equation:
[tex]\begin{gathered} C=\frac{n_{AgNO_3}}{V_{\text{solution}}} \\ n_{AgNO_3}=C\cdot V_{\text{solution}} \end{gathered}[/tex]So we have:
[tex]m_{AgCl}=M_{AgCl}\cdot C\cdot V_{\text{solution}}[/tex]The molar mass of AgCl is the value given in parenthesis and the concentration and volume were given at the beginning:
[tex]\begin{gathered} M_{AgCl}=143.32g/mol \\ V=29.48mL=29.48\times10^{-3}L \\ C=1.524M=1.524mol/L \end{gathered}[/tex]Putting them altogether, we have:
[tex]\begin{gathered} m_{AgCl}=143.32g/mol\cdot29.48\times10^{-3}L\cdot1.524mol/L \\ m_{AgCl}=143.32\cdot29.48\times10^{-3}\cdot1.524\cdot g \\ m_{AgCl}=6.439012\ldots g\approx6.439g \end{gathered}[/tex]So, the mass that we should obtain assuming all reacts is approximately 6.439 g.
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