Percent yield/excess reactant problem

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In summary, the problem involves the reaction of N2 and H2 to form NH3 with 100% yield. There are 2.0 moles of excess reactants left over. The solution can be found by setting up equations or using 'smart' trial and error. By setting up equations, we can solve for the initial amounts of reactants. By using 'smart' trial and error, we can first balance the equation and then determine the initial amounts of reactants needed to be left with 2 moles of excess reactants.
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Lori

Homework Statement



N2 reacts with H2 to form NH3

assume equal moles of N2 and H2 in container, and react to form NH3 with 100% yield. there are 2.0 moles of excess reactants left over. How many moles of each reactant was originally present?

Homework Equations


AY/TY = % yield

The Attempt at a Solution


I know that the answer is 3 moles after much trial and error. Cause if i had started with 3 moles of each reactant, then i used 1 mole of N2. Which means 2 moles in excess. How could i have shorten this problem without doing trial error?
 
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  • #2
Lori said:

Homework Statement



N2 reacts with H2 to form NH3

assume equal moles of N2 and H2 in container, and react to form NH3 with 100% yield. there are 2.0 moles of excess reactants left over. How many moles of each reactant was originally present?

Homework Equations


AY/TY = % yield

The Attempt at a Solution


I know that the answer is 3 moles after much trial and error. Cause if i had started with 3 moles of each reactant, then i used 1 mole of N2. Which means 2 moles in excess. How could i have shorten this problem without doing trial error?

Hi Lori! :oldsmile:

We can set up equations, or we can do 'smart' trial and error.

With equations
We have 2 unknowns: the initial amount of reactants, and the amount of NH3 formed.
Let's call them x and y.
Then the equation becomes: x N2 + x H2 → y NH3 + 2 N2
Can we make a set of equations from that and solve it?

With 'smart' trial and error
Let's first balance the equation:
? N2 + ? H2 → ? NH3 + ? N2
Can we find the question marks such that the equation is balanced?
After that we need to figure out what to do to ensure we're left with 2 moles of initial reactant.
 
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Related to Percent yield/excess reactant problem

1. What is percent yield in a chemical reaction?

Percent yield is a measurement of the efficiency of a chemical reaction in producing the desired product. It is calculated by dividing the actual yield (the amount of product obtained in the reaction) by the theoretical yield (the amount of product that should have been obtained based on stoichiometric calculations) and multiplying by 100%. This gives a percentage that represents the proportion of the expected product that was actually obtained.

2. How is percent yield calculated?

To calculate percent yield, divide the actual yield by the theoretical yield and multiply by 100%. The actual yield can be determined by measuring the amount of product obtained in the reaction. The theoretical yield is calculated using stoichiometric calculations based on the balanced chemical equation.

3. What is an excess reactant?

An excess reactant is a reactant that is present in a greater amount than is required for the reaction to reach completion. This means that some of the excess reactant will remain unreacted after the reaction is complete.

4. How do you determine the excess reactant in a chemical reaction?

To determine the excess reactant in a chemical reaction, first calculate the theoretical yield of both reactants based on the balanced chemical equation. Then, compare the actual yield of each reactant to their respective theoretical yields. The reactant with a greater difference between the actual and theoretical yield is the excess reactant.

5. How does the concept of percent yield and excess reactant relate to the efficiency of a reaction?

The concept of percent yield and excess reactant are both related to the efficiency of a reaction. Percent yield measures the efficiency of a reaction in producing the desired product, while the presence of an excess reactant indicates that the reaction was not completely efficient in using all of the reactants to produce the desired product. In other words, a higher percent yield and a lower amount of excess reactant both indicate a more efficient reaction.

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