Consider the reaction of 19.0 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 29.0 g of solid metal is present. Calculate the mass of each metal in the 29.0-g mixture

First, let's write the balanced chemical equation for the reaction between zinc and silver nitrite: \( Zn + 2AgNO₂ → 2Ag + Zn(NO₂)₂ \)

In this reaction, we are given excess silver nitrite and 19.0 g of zinc. So, zinc is the limiting reactant.

Now, let's calculate the theoretical yield of the product mixture by determining the moles of zinc and using the stoichiometry of the balanced equation. The molar mass of zinc is 65.38 g/mol, so the moles of zinc used are: moles Zn = \( \frac{19.0 g}{65.38 \frac{g}{mol}} \) = 0.2907 mol Now, let's use the stoichiometry of the balanced equation to find the theoretical yield of silver and zinc nitrite. For every mole of zinc reacted, 2 moles of silver are produced. moles Ag = 2 * moles Zn = 2 * 0.2907 mol = 0.5814 mol The molar mass of silver is 107.87 g/mol, so the mass of silver produced is: mass Ag = moles Ag * molar mass Ag = 0.5814 mol * 107.87 g/mol = 62.70 g Since the actual mass of the product mixture is 29.0 g and the mass of silver produced is more than the mass of the product mixture, this means not all the zinc has reacted.

To find the mass of each metal in the 29.0-g mixture, we will subtract the mass of unreacted zinc from the mass of the product mixture to get the mass of silver present and use the mass of silver to find the mass of reacted zinc. Let x represent the mass of unreacted zinc, so the mass of silver in the mixture is (29.0 - x) g. Now let's use the molar ratio between zinc and silver: \( \frac{x}{65.38 \frac{g}{mol}} = \frac{29.0 - x}{107.87 \frac{g}{mol}} \) Solving for x, we get: x = 10.0 g (mass of unreacted zinc) Now, we can calculate the mass of silver and reacted zinc in the 29.0-g mixture: mass Ag = 29.0 g (total mass) - 10.0 g (unreacted zinc) = 19.0 g (silver) mass reacted Zn = 19.0 g (initial zinc) - 10.0 g (unreacted zinc) = 9.0 g (reacted zinc) So, in the 29.0-g mixture, 19.0 g is silver, and 9.0 g is reacted zinc.