Consider the reaction of \(19.0 \mathrm{~g}\) 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 \(29.0 \mathrm{~g}\) of solid metal is present. Calculate the mass of each metal in the \(29.0-\mathrm{g}\) mixture.

The balanced chemical equation for the reaction between zinc (Zn) and silver nitrite (AgNO2) is: \[2 \mathrm{AgNO_2} + \mathrm{Zn} \rightarrow 2 \mathrm{Ag} + \mathrm{Zn(NO_2)_2}\]

To determine the initial number of moles of zinc, we will first find the molar mass of Zn. The atomic mass of Zn is approximately 65.38 g/mol. Therefore, the initial number of moles of Zn is: \[\text{moles of Zn} = \frac{\text{mass of Zn}}{\text{molar mass of Zn}}\] \[\text{moles of Zn} = \frac{19.0 \mathrm{~g}}{65.38 \mathrm{~g/mol}}\] \[\text{moles of Zn} \approx 0.2906 \mathrm{~mol}\]

If not all of the zinc has reacted, then the final mass of zinc in the \(29.0 \mathrm{~g}\) metal mixture will be the initial mass of zinc minus the mass of zinc that reacted. Let's denote the mass of reacted zinc as \(x\). Then the final mass of zinc will be: \[\text{final mass of Zn} = 19.0 \mathrm{~g} - x\] The moles of Zn that reacted can be found using the stoichiometry of the balanced equation: \[\text{moles of reacted Zn} = \frac{x}{65.38 \mathrm{~g/mol}}\] Since 2 moles of silver are produced for every mole of zinc that reacted: \[\text{moles of produced Ag} = 2 \times \frac{x}{65.38 \mathrm{~g/mol}}\] \[\text{mass of produced Ag} = \text{moles of produced Ag} \times \text{molar mass of Ag}\] \[\text{mass of produced Ag} = 2 \times \frac{x}{65.38 \mathrm{~g/mol}} \times 107.87 \mathrm{~g/mol}\] Using the conservation of mass: \[\text{mass of produced Ag} + \text{final mass of Zn} = 29.0 \mathrm{~g}\] \[2 \times \frac{x}{65.38 \mathrm{~g/mol}} \times 107.87 \mathrm{~g/mol} + 19.0 \mathrm{~g} - x = 29.0 \mathrm{~g}\] Now, we need to solve the equation for x value, the mass of the reacted Zn.

Solving for x, we get: \[x \approx 9.613\mathrm{~g}\] Now we can calculate the final mass of Zn: \[\text{final mass of Zn} = 19.0 \mathrm{~g} - 9.613 \mathrm{~g} \approx 9.387 \mathrm{~g}\] To find the mass of silver in the mixture, substitute the value of x back into the equation for mass of produced Ag: \[\text{mass of produced Ag} = 2 \times \frac{9.613 \mathrm{~g}}{65.38 \mathrm{~g/mol}} \times 107.87 \mathrm{~g/mol} \approx 19.613 \mathrm{~g}\] So, in the \(29.0-\mathrm{g}\) mixture, there are approximately \(9.387\, \mathrm{g}\) of zinc and \(19.613\, \mathrm{g}\) of silver.