Respuesta :
Answer:
4.873g of AgCl will precipitate.
Explanation:
1st) It is necessary to calculate the moles of magnesium chloride contained in 39.6mL of 0.428M solution:
[tex]\begin{gathered} 1000mL-0.428moles \\ 39.6mL-x=\frac{39.6mL*0.428moles}{1000mL} \\ x=0.017moles \end{gathered}[/tex]Now we know that there are 0.017moles of magnesium chloride in the solution.
2nd) According to the stoichiometry of the reaction, from 1 mole of magnesium chloride, 2 moles AgCl precipitate, so we can use a mathematical rule of three and the 0.017moles of MgCl2 to calculate the moles of AgCl that will precipitate:
[tex]\begin{gathered} 1moleMgCl_2-2molesAgCl \\ 0.017moleMgCl_2-x=\frac{0.017moleMgCl_2*2molesAgCl}{1moleMgCl_2} \\ x=0.034molesAgCl \\ \end{gathered}[/tex]Now we know that 0.034 moles of AgCl are produced.
3rd) Finally, we can calculate the grams of silver chloride by using a mathematical rule of three with the molar mass of AgCl (143.32g/mol) and the 0.034moles:
[tex]\begin{gathered} 1mol-143.32g \\ 0.034moles-x=\frac{0.034moles*143.32g}{1mol} \\ x=4.873g \end{gathered}[/tex]So, 4.873g of AgCl will precipitate.
