Why was Rutherford's Gold Foil Experiment so important?
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Taking into account the reaction stoichiometry and definition of limiting reactant, AgNO₃ will be the limiting reagent.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Cu: 1 mole
- AgNO₃: 2 moles
- Ag: 2 moles
- Cu(NO₃)₂: 1 mole
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
<h3>Limiting reagent in this case</h3>To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Cu reacts with 2 moles of AgNO₃, 1.8 moles of Cu reacts with how many moles of AgNO₃?
<u><em>amount of moles of AgNO₃= 3.6 moles</em></u>
But 3.6 moles of AgNO₃ are not available, 2 moles are available. Since you have less moles than you need to react with 1.8 moles of Cu, AgNO₃ will be the limiting reagent.
<h3>Summary</h3>In summary, AgNO₃ will be the limiting reagent.
Learn more about the reaction stoichiometry:
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Answer:
36.37% is the percent yield of the reaction.
Explanation:
1)0.650 L nitrogen gas , at 295 K and 1.01 bar.
Let the moles of nitrogen gas be n.
Pressure of the gas ,P= 1.01 bar = 0.9967 atm (1 bar = 0.9869 atm)
Temperature of the gas = T = 295 K
Volume of the gas = V = 0.650 L
Using an ideal gas equation:
2) Moles of ammonia gas=
Moles of oxygen gas =
According to reaction ,3 mol of oxygen reacts with 4 mol of ammonia.
Then,0.1101 mol of oxygen will react with:
of ammonia.
Hence, oxygen gas is in limiting amount and act as limiting reagent.
3) Theoretical yield of nitrogen gas :
According to reaction, 3 mol of oxygen gas gives 2 moles of nitrogen gas.
Then 0.1101 mol of oxygen will give:
of nitrogen.
Theoretical yield of nitrogen gas = 0.0734 mol
Experimental yield of nitrogen as calculated in part (1) = 0.0267 mol
Percentage yield:
Percentage yield of the reaction:
36.37% is the percent yield of the reaction.