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Chapter 4: Problem 125

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-g mixture.

Short Answer

Expert verified
In the 29.0 g mixture, there are \(3.319\,\text{g}\) of unreacted zinc and \(25.681\,\text{g}\) of silver.

Step by step solution

01

Write the balanced chemical equation for the reaction

For the given reaction, zinc (Zn) reacts with silver nitrite (AgNO₂) to produce silver (Ag) and zinc nitrite (Zn(NO₂)₂). So, the balanced chemical equation would be: \[Zn + 2AgNO_2 \rightarrow 2Ag + Zn(NO_2)_2\]
02

Find the moles of zinc given

We are given 19.0 g of zinc. To find the number of moles of zinc, we will use the molar mass of zinc, which is 65.38 g/mol. So, the moles of zinc will be: \[\frac{19.0\,\text{g}}{65.38\,\text{g/mol}} = 0.2907\,\text{mol} \]
03

Determine the mass of silver produced

From the balanced chemical equation in Step 1, we see that 1 mole of zinc reacts with 2 moles of silver nitrite to produce 2 moles of silver. So, the moles of silver produced will be equal to the moles of zinc. Using the moles of zinc calculated in Step 2, and the molar mass of silver (107.87 g/mol), we can find the mass of silver produced: \[0.2907\,\text{mol} \times 107.87\,\text{g/mol} = 31.36\,\text{g}\] However, the total mass of solid metal is given to be 29.0 g, which means the reaction did not go to completion and not all the zinc reacted.
04

Calculate the mass of unreacted zinc

If all the zinc had reacted, we would have had 31.36 g of silver. But since we only have 29.0 g of solid metal mixture, we need to find out how much of the calculated silver was actually produced and subtract that from the total mass to find the mass of unreacted zinc. Let x be the mass of the unreacted zinc, then the mass of actually produced silver is (29.0 - x) g. From the stoichiometry of the reaction, we can write: \[\frac{29.0-x\,\text{g}}{107.87\,\text{g/mol}} = \frac{0.2907\,\text{mol}}{2}\] Solving for x, we get: \[x = 19.0 - \left( \frac{0.2907\,\text{mol}}{2} \times 107.87\,\text{g/mol} \right) = 3.319\,\text{g}\]
05

Find the mass of each metal in the mixture

Now, we know the mass of unreacted zinc (3.319 g), and we can calculate the mass of silver in the mixture by subtracting the mass of unreacted zinc from the total mass (29.0 g): \[mass\,of\,Ag = 29.0\,\text{g} - 3.319\,\text{g} = 25.681\,\text{g}\] So, in the 29.0 g mixture, there are 3.319 g of unreacted zinc and 25.681 g of silver.

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Most popular questions from this chapter

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