Gas Laws Fact Sheet
| Ideal gas law | [tex]PV=nRT[/tex] |
Answer (2)
The Ideal Gas Law, represented by the equation P V = n RT , connects pressure, volume, temperature, and moles of a gas. It is significant in understanding gas behaviors and applies to various practical situations, like filling balloons or predicting gas behaviors in engines. This law combines Boyle's, Charles's, and Avogadro's Laws into a single equation for comprehensive gas calculations.
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Determine the moles of lead(II) nitrate ( P b ( N O 3 ) 2 ) and aluminum (Al).
Identify lead(II) nitrate as the limiting reactant.
Use the stoichiometry of the balanced equation to find the moles of lead (Pb) produced.
Calculate the theoretical yield of lead: 5.41 g .
Explanation
Balanced Equation First, we need to identify the limiting reactant in the given reaction. The balanced chemical equation is: 2 A l ( s ) + 3 P b ( N O 3 ) 2 ( a q ) → 3 P b ( s ) + 2 A l ( N O 3 ) 3 ( a q )
Molar Masses We are given 8.65 grams of lead(II) nitrate ( P b ( N O 3 ) 2 ) and 2.50 grams of aluminum (Al). To determine the limiting reactant, we need to calculate the number of moles of each reactant. The molar mass of P b ( N O 3 ) 2 is 331.20 g/mol, the molar mass of Al is 26.98 g/mol, and the molar mass of Pb is 207.2 g/mol.
Moles of Lead(II) Nitrate Calculate the moles of P b ( N O 3 ) 2 : m o l es = m o l a r ma ss ma ss = 331.20 g / m o l 8.65 g = 0.02612 m o l
Moles of Aluminum Calculate the moles of Al: m o l es = m o l a r ma ss ma ss = 26.98 g / m o l 2.50 g = 0.09266 m o l
Limiting Reactant Now, we need to determine the limiting reactant by comparing the mole ratio of the reactants to the stoichiometric ratio in the balanced equation. From the balanced equation, the mole ratio of Al to P b ( N O 3 ) 2 is 2:3.
Identifying Limiting Reactant To determine the limiting reactant, we can calculate how much P b ( N O 3 ) 2 is required to react completely with the given amount of Al: m o l es o f P b ( N O 3 ) 2 re q u i re d = 2 3 × m o l es o f A l = 2 3 × 0.09266 m o l = 0.13899 m o l Since we only have 0.02612 mol of P b ( N O 3 ) 2 , it is the limiting reactant.
Theoretical Yield of Lead Now we can calculate the theoretical yield of lead (Pb) based on the limiting reactant. From the balanced equation, 3 moles of Pb are produced for every 3 moles of P b ( N O 3 ) 2 . Therefore, the mole ratio of P b ( N O 3 ) 2 to Pb is 1:1.
Moles of Lead Produced Calculate the moles of Pb produced: m o l es o f P b = m o l es o f P b ( N O 3 ) 2 = 0.02612 m o l
Mass of Lead Produced Calculate the theoretical yield of Pb: ma ss = m o l es × m o l a r ma ss = 0.02612 m o l × 207.2 g / m o l = 5.41 g
Final Answer Therefore, the theoretical yield of solid lead is 5.41 g.
Examples
In the field of environmental science, determining the theoretical yield of a precipitate formed during water treatment is crucial. For instance, when removing heavy metals from contaminated water, scientists calculate the amount of precipitating agent needed and the expected yield of the metal precipitate to optimize the treatment process and ensure regulatory compliance. This calculation relies on stoichiometry, similar to the problem we solved, ensuring efficient and effective removal of pollutants.
