Questions: Consider the reaction of 21.0 g of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and 32.0 g of solid metal is present. Calculate the mass of each metal in the 32.0 g mixture. Mass of Zn= g Mass of Ag= g

Solution Steps

The reaction between zinc (Zn) and silver nitrite (AgNO\(_2\)) can be represented as:

\[ \text{Zn} + 2\text{AgNO}_2 \rightarrow 2\text{Ag} + \text{Zn(NO}_2\text{)}_2 \]

First, calculate the moles of zinc initially present using its molar mass (65.38 g/mol):

\[ \text{Moles of Zn} = \frac{21.0 \, \text{g}}{65.38 \, \text{g/mol}} = 0.3212 \, \text{mol} \]

Let \( x \) be the mass of unreacted zinc. The mass of silver in the mixture will then be \( 32.0 \, \text{g} - x \).

The moles of zinc that reacted will produce an equivalent amount of moles of silver, as per the stoichiometry of the reaction. Therefore, the moles of silver produced is:

\[ \text{Moles of Ag} = 0.3212 \, \text{mol} - \frac{x}{65.38 \, \text{g/mol}} \]

The mass of silver produced is:

\[ \text{Mass of Ag} = \left(0.3212 - \frac{x}{65.38}\right) \times 107.87 \, \text{g/mol} \]

Since the total mass of the mixture is 32.0 g:

\[ x + \left(0.3212 - \frac{x}{65.38}\right) \times 107.87 = 32.0 \]

Simplify and solve for \( x \):

\[ x + 34.62 - \frac{107.87x}{65.38} = 32.0 \]

\[ x - \frac{107.87x}{65.38} = -2.62 \]

\[ x \left(1 - \frac{107.87}{65.38}\right) = -2.62 \]

\[ x \left(\frac{-42.49}{65.38}\right) = -2.62 \]

\[ x = \frac{-2.62 \times 65.38}{-42.49} = 4.032 \, \text{g} \]

The mass of silver is:

\[ 32.0 \, \text{g} - 4.032 \, \text{g} = 27.968 \, \text{g} \]

Final Answer