Stoichiometry Reactions: Ammonia
What is stoichiometry?
Excess Reactant: the reactant that has higher yield than the limiting reactant
Limitng Reactant: the reactant that has the lowest theoretical yield due to its lack of quantity. This is the reactant that would be exhausted the quickest
Molar mass: the amount of grams it would weigh for the one mole of an object
Mole: a standard scientific unit for measuring large quantities of very small entities such as atoms, molecules, or other specified particles
Product: the substances that result form the recombination of atoms
Reactant: the substance that takes part in and undergoes change during a reaction
Stoichiometry: the measurement and calculation of the amounts of reactants and products in chemical equations
Theoretical yield: how much something can be produced when given value of something else; equivalent to the limiting reactant's yield
Percent yield: displays how much of a reaction was produced versus the predicted, or theorized amount. It is displayed in percentages and it is actual yield/ theoretical yield
Today's Reaction: Nitrogen and Hydrogen yields Ammonia
How to Solve Stoichiometry Problems
- N2 + 3H2 -> 2NH3 H A
After balancing the equation out, we can now start looking at the stoichiometry.
- Balancing equation: you have to add coefficients in the equation to make sure that there remains the same number of substances remain. This is to fulfill the law of conservation of mass. The equation above is already balanced.
- Molar mass: molar mass is obtained from finding its each respective molar masses of every element listed in the substance and multiplying by its subscripts and adding the product of all the elements in the substance. You do not multiply by its coefficient. In this case, 2*(14.007)= 28.014 g for the nitrogen, 2*(1.008)= 2.016 g for hydrogen, and 14.007+3*(1.008)= 17.031 g for ammonia.
- Mole to mole conversions: in order for mole to mole conversion, you must understand that the coefficients on the equation actually tells you the ratio of the moles between substances in the equation. Because of this, the product of the reactant A's given mole and reactant B's coefficient should equal to the product of the reactant B's given mole and reactant B's coefficient. Much more simply, multiply the given mole of reactant A by the ratio of the two substances, where reactant B's coefficient on the numerator and the reactant A coefficient on the denominator. For this example, I will use 10.04 moles of nitrogen to find the number of moles of ammonia. I will multiply 10.04 by 2 as the ratio between the ammonia and the nitrogen is 2:1. As a result, I get 20.08 moles of ammonia.
- Mass to mass conversions: mass to mass is one more step than mole to mole conversions. First you have to find how many moles of the given substance you have when given its mass. This is achieved by diving the given mass by its molar mass. Then, you must multiply by its ratio in order to find out how many moles of substance B it can produce (this is mole to mole conversion). Then, additionally, you have to multiply the amount of mole of reactant B by its molar mass to get the final mass. This is to actually calculate how much mass it is when its not 1 mole of that final product. For this example, I will use 12.1 g of nitrogen and convert it to mass of ammonia. I will divide 12.1 by 28.014, multiply the quotient by 2, and multiply the product by 17.031. The solution comes out to 14.71229385 and this is reduced to 14.7 g as it was rounded to least significant figure in this arithmetic.
- Limiting and excess reactant: because limiting reactant limits the amount of product it can be produced due to its lack of quantity, the lowest yield out of the two reactants is the limiting reactant. Conversely, the other reactant that does not have the lowest yield is the excess reactant. For this, you must solve mass to mass conversions for every single reactant given and compare the results. In this example, I will be using 12.1 g for nitrogen and 20.15 g of hydrogen to see how much mass they can produce ammonia based on their given mass. For nitrogen, we already have calculated the mass of ammonia, which was 14.7 g. For hydrogen, you divide 20.15 by 2.015, multiply the quotient by 2/3 (ratio), and multiply the product by 17.031. This produces 113.4836806, or 113.5 g of ammonia when given 20.15 g hydrogen. The limiting reactant in this case is nitrogen as it only allows to create 14.7 g than 113.5 g of hydrogen's; the excess reactant is the hydrogen.
- Theoretical yield: the theoretical yield is the yield of the limiting reactant, which is 14.7 g.
- Percent yield: its the quotient of the actual amount and the theoretical amount. In this case, assume that 6.5 g of ammonia has been produced with the same conditions as before when calculating the theoretical yield. 6.5 divided by 14.7 is approximately .44 and you have to multiply by 100 to get the percentage. Thus, the percent yield is 44%.
Research About the Reaction: Ammonia
Images and Sources for Additional Help
this is a visual representation of the many stoichiometry conversions that take place.
this one takes one further step and shows the processes of getting formula units of A and B.