Question

If x g weight of ${\text{AgCl}}$ will be precipitated when a solution containing 4.77 g ${\text{NaCl}}$ is added to a solution of 5.77 g of ${\text{AgN}}{{\text{O}}_{\text{3}}}$, then find the value of x.
A. 143.5
B. 142.5
C. 147.5
D. None of the above

Answer 1

Write the balanced chemical reaction for the production of ${\text{AgCl}}$ using ${\text{NaCl}}$ and ${\text{AgN}}{{\text{O}}_{\text{3}}}$. From the given masses of the reactants calculate the moles of both the reactants. Determine the limiting reagent. Using the moles of the limiting reagent calculate the moles of the product ${\text{AgCl}}$. Finally, convert the moles product to the mass of the product.

Complete Step by step answer: We have given that ${\text{AgCl}}$ will be precipitated when ${\text{NaCl}}$ reacts with ${\text{AgN}}{{\text{O}}_{\text{3}}}$ .
So first we will write a balanced chemical reaction.
${\text{NaCl + AgN}}{{\text{O}}_{\text{3}}} \to {\text{ AgCl }} \downarrow + {\text{ NaN}}{{\text{O}}_{\text{3}}}$
Now, using the given masses of ${\text{NaCl}}$ and ${\text{AgN}}{{\text{O}}_{\text{3}}}$ we will calculate the moles of both the reactants.
${\text{moles = }}\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$
Mass of ${\text{NaCl}}$ = 4.77 g
Molar mass of ${\text{NaCl}}$ = 58.44 g/mol
So, ${\text{moles of NaCl = }}\dfrac{{{\text{4}}{\text{.77 g}}}}{{58.44{\text{ g/mol}}}} = {\text{0}}{\text{.0816 mole}}$
Similarly, we can calculate the moles of ${\text{AgN}}{{\text{O}}_{\text{3}}}$ as follows:
Mass of ${\text{AgN}}{{\text{O}}_{\text{3}}}$ = 5.77 g
Molar mass of ${\text{AgN}}{{\text{O}}_{\text{3}}}$ = 169.87 g/mol
So, ${\text{moles of AgN}}{{\text{O}}_{\text{3}}}{\text{ = }}\dfrac{{{\text{5}}{\text{.77 g}}}}{{{\text{169}}{\text{.87 g/mol}}}} = {\text{0}}{\text{.03396 mole}}$
Form the balanced chemical equation we can say that the mole ratio of ${\text{NaCl}}$:${\text{AgN}}{{\text{O}}_{\text{3}}}$:${\text{AgCl}}$ is 1:1:1
As we have 0.0816 mol of ${\text{NaCl}}$ and 0.03396 mol of ${\text{AgN}}{{\text{O}}_{\text{3}}}$ so we can say 0.03396 mol of ${\text{AgN}}{{\text{O}}_{\text{3}}}$ will react with 0.03396 mol of ${\text{NaCl}}$ and will give 0.03396 mol of ${\text{AgCl}}$
Here ${\text{AgN}}{{\text{O}}_{\text{3}}}$ is the limiting reagent so we have determined the moles of product from moles of limiting reagent.
Now, we have moles of ${\text{AgCl}}$. Now we have to convert these moles of ${\text{AgCl}}$ to mass using its molar mass as follows:
Molar mass of ${\text{AgCl}}$ = 143.5 g/mol
$\Rightarrow {\text{Mass of AgCl}} = {\text{moles \times molar mass}}$
$\Rightarrow {\text{Mass of AgCl}} = 0.03396{\text{ g }} \times {\text{143}}{\text{.5 g/mol}}$
$\Rightarrow {\text{Mass of AgCl}} = 4.87{\text{ g}}$
Thus, we can say that 4.87 g of ${\text{AgCl}}$will be precipitated out when a solution containing 4.77 g ${\text{NaCl}}$ is added to a solution of 5.77 g of${\text{AgN}}{{\text{O}}_{\text{3}}}$.

Thus the correct option is (D) None of the above.

Note: Limiting reagent is the reagent which completely reacts in the reaction. So, the moles of limiting reagent and stoichiometric ratio of limiting reagent with product determine the amount of product. In mole concept people generally make mistakes in finding the exact solution due to unbalanced chemical equations.