Question

Silver nitrate reacts with sodium chloride to form sodium nitrate and silver chloride. It is found that 6.8 g of silver nitrate reacts with 2.4 g of sodium chloride and forms 3.4 g of sodium nitrate. Applying the law of conservation of mass, calculate the weight of silver chloride formed.

Answer 1

Reaction of Silver nitrate with Sodium chloride:

• Silver nitrate (${\mathrm{AgNO}}_3$) reacts with Sodium chloride ($\mathrm{NaCl}$) to form Silver chloride ($\mathrm{AgCl}$) and Sodium nitrate (${\mathrm{NaNO}}_3$) as products.
• The reaction is given below:

$$\begin{gathered} {\mathrm{AgNO}}_3\left(\mathrm{s}\right)+\mathrm{NaCl}\left(\mathrm{s}\right)\to ️\mathrm{AgCl}\left(\mathrm{s}\right)+{\mathrm{NaNO}}_3\left(\mathrm{s}\right) \\ 6.8\mathrm{g}+2.4\mathrm{g}\to \mathrm{X}+3.4\mathrm{g} \end{gathered}$$

• Let the unknown mass of Sodium chloride be X.
• By the law of conservation of mass,

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{reactants}=\mathrm{Mass}\mathrm{of}\mathrm{products} \\ \Rightarrow 6.8+2.4=\mathrm{X}+3.4 \\ \Rightarrow 9.2=\mathrm{X}+3.4 \\ \Rightarrow \mathrm{X}=9.2-3.4 \\ \Rightarrow \mathrm{X}=5.8\mathrm{g} \end{gathered}$$

• Hence, the weight of silver chloride formed is 5.8g.