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Chapter 6: Problem 127

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 9.0 g of zinc remaining and 20.0 g of silver produced.

Step by step solution

01

Write the balanced chemical equation for the reaction

The reaction between zinc metal and silver nitrite can be represented as: \[Zn + 2AgNO_2 \rightarrow 2Ag + Zn(NO_2)_2\]
02

Calculate the moles of zinc initially present

To calculate the moles of zinc initially present in the reaction, use the given mass and the molar mass of zinc (65.38 g/mol): Moles of zinc = \(\frac{mass}{molar\ mass}\) Moles of zinc = \(\frac{19.0 \ g}{65.38 \ g/mol}\) = 0.2906 mol
03

Determine the maximum amount of silver that can be produced

According to the balanced chemical equation, 1 mole of zinc can produce 2 moles of silver. Therefore, we can calculate the maximum moles of silver that can be produced: Maximum moles of silver = moles of zinc × \(2\) Maximum moles of silver = 0.2906 mol × \(2\) = 0.5812 mol Now, we can find the mass of the maximum amount of silver: Mass of maximum silver = moles × molar mass Mass of maximum silver = 0.5812 mol × \(107.87 \ g/mol\) = 62.7 g
04

Determine the mass of unreacted zinc

Since the reaction stopped before all the zinc was consumed, some zinc remains unreacted. We know that the total mass of solid metal at the end of the reaction is 29.0 g, so we can find the mass of unreacted zinc: Mass of unreacted zinc = Total mass of solid metal - Mass of maximum silver Mass of unreacted zinc = \(29.0 \ g - 62.7 \ g\) = -33.7 g The negative value indicates that there is less silver produced than the maximum possible amount since not all the zinc was consumed, and we should recalculate for this case.
05

Determine the actual mass of silver produced and zinc remaining

Assume 'x' grams of zinc reacted. Then, according to the stoichiometry, 2 times the mass of silver is produced: Mass of silver = \(2x\) Mass of remaining zinc = \(19.0 \ g - x\) The total mass of solid metal in the mixture is given as 29.0 g: Total mass of solid metal = Mass of silver + Mass of remaining zinc 29.0 g = \(2x + 19.0 \ g - x\) Now, solving for 'x': x = 10.0 g So, 10.0 g of zinc reacted, and the mass of the remaining zinc is: Mass of remaining zinc = \(19.0 \ g - 10.0 \ g\) = 9.0 g The mass of silver produced is: Mass of silver = \(2 × 10.0 g\) = 20.0 g
06

Conclusion

In the 29.0 g mixture, there are 9.0 g of zinc remaining and 20.0 g of silver produced.

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