Question

How many grams of silver chloride will be precipitated by adding sufficient silver nitrate to reactwith 39.6 mL of 0.428 M magnesium chloride?2 AgNO3 (aq) + MgCl₂ (aq) --> 2 AgCl (s) + Mg(NO3)2 (aq)

Answer 1

4.873g of AgCl will precipitate.

Step-by-step explanation:

1st) It is necessary to calculate the moles of magnesium chloride contained in 39.6mL of 0.428M solution:

`\begin{gathered} 1000mL-0.428moles \\ 39.6mL-x=(39.6mL*0.428moles)/(1000mL) \\ x=0.017moles \end{gathered}`

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:

`\begin{gathered} 1moleMgCl_2-2molesAgCl \\ 0.017moleMgCl_2-x=(0.017moleMgCl_2*2molesAgCl)/(1moleMgCl_2) \\ x=0.034molesAgCl \\ \end{gathered}`

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:

`\begin{gathered} 1mol-143.32g \\ 0.034moles-x=(0.034moles*143.32g)/(1mol) \\ x=4.873g \end{gathered}`

So, 4.873g of AgCl will precipitate.