All AP Chemistry Resources
Example Questions
Example Question #32 : Stoichiometry
For the molecular formula, , how many grams of oxygen are in 50.0g of this molecule?
The first step is to determine the molecular weight of the molecule ethanol, :
The molecular weight serves as a conversion factor to convert grams to moles as implied by its units . So, given that we are dealing with 50 grams of ethanol, the moles of 50 grams can be calculated using the calculated molecular weight as seen below in which the grams cancel out leaving the units in moles:
Based on the molecular formula, we can see that for every mole of the compound , there is 1 mole of oxygen. Based on our calculations, 50 grams of is 1.09 moles of the compound. Because there is a 1:1 molar ratio of oxygen to every ethanol molecule, that is, 1 mole of oxygen in every mole of our molecule, we have 1.09 moles of oxygen in 50 grams of our compound . To convert 1.09 moles of oxygen to grams of oxygen, we perform the following equation based on the atomic weight of oxygen, which from the periodic table we know is :
Example Question #6 : Other Stoichiometric Calculations
For the molecular formula, , how many grams of oxygen are in 45.0 grams of this molecule?
The first step is to determine the molecular weight of the molecule:
The molecular weight serves as a conversion factor to convert grams to moles as implied its units . So, given that we are dealing with 45.0 grams, the moles of 45.0 grams can be calculated using the calculated molecular weight as seen below in which the grams cancel out leaving the units in moles:
Based on the molecular formula, we can see that for every mole of the compound , there are 2 moles of oxygen. Based on our calculations 45.0 grams of is 0.506 moles of the compound. Because there is a 2:1 mole ratio of oxygen to every molecule, that is, 2 mole of oxygen in every mole of our molecule, we have double the moles of oxygen in our compound as calculated below:
To convert 1.011 moles of oxygen to grams of oxygen, we perform the following equation based on the atomic weight of oxygen, which is :
Example Question #451 : Ap Chemistry
Consider the reaction of baking soda and vinegar in an aqueous solution:
If we begin with 17g of and excess , what is the volume of the gas released?
Assume that all gases behave as ideal gases at atmospheric pressure. The temperature is 298K.
We begin with 17 g of, so our first step is to calculate the molecular weight of .
Now with the molecular weight, we will calculate the number of moles within 17 g of .
Now we will use this to calculate the number of moles of gas produced. In this reaction, the only gas produced is carbon dioxide.
An ideal gas at 1 atmosphere and 298 K has the molar volume of .
Example Question #35 : Stoichiometry
What volume of 0.5M NaOH is necessary to neutralize 15mL of 1.0M nitrous acid?
10mL
1L
30L
30mL
30mL
Both the acid and the base provide one equivalent of the necessary ions (H+ and OH–, respectively) and therefore you can use the equation M1V1 = M2V2.
(1)(15) = (0.5)V2
V2 = 30mL
Example Question #2 : Stoichiometry With Reactions
What volume of 0.5M Mg(OH)2 is necessary to neutralize 15mL of 1.0M nitrous acid?
1L
15mL
15L
10mL
30mL
30mL
This problem is tricky because you must realize that Mg(OH)2 will release 2 molecules of OH– whenever one dissolves; thus, you only need half the amount of Mg(OH)2 to neutralize all of the nitrous acid, which only gives 1 equivalent.
M1V1 = M2V2
Plugging in the values, you get 30mL, but in reality you only need half of that amount.
Example Question #1 : Stoichiometry With Reactions
Magnesium will combine with oxygen gas to form magnesium oxide according to the balanced equation below.
If 65g of MgO are created in the reaction, how many grams of magnesium are used in the reaction?
In order to determine how many grams of magnesium are used in the reaction, we can use the percentage by mass of magnesium in MgO in order to find out how much magnesium is in 65 grams of MgO. The molar mass of MgO is 40.3 grams. The molar mass of magnesium is 24.3 grams. Dividing 24.3 by 40.3 reveals that magnesium makes up 60% of the mass of MgO.
Multiplying 65 grams by 0.603 reveals that there are 39.2 grams of magnesium in 65 grams of MgO.
An alternative means of solving the problem would be to convert grams of MgO to moles. The ratio of MgO to Mg in the given chemical equation is 2:2, thus the number of moles of MgO formed is equal to the number of moles of Mg reacted. Once this number is found, it can be converted back to grams using the molar mass of magnesium.
Example Question #4 : Stoichiometry With Reactions
Consider the following chemical reaction:
How many moles of will be produced if 75.0mL of water is produced?
There is not enough information to answer this question
4.17mol
None of the available answers
1.56 moles
1.39mol
1.39mol
This problem is solved based on stoichiometry. Using the molecular weights and ratios in the given reaction, we can solve for the amount of produced.
Example Question #5 : Stoichiometry With Reactions
The formation of ammonia is given by the following equation:
Assuming you started with 20g of hydrogen gas, how much nitrogen gas is necessary to react the hydrogen gas?
60g
40g
6.67g
93.33g
93.33g
We can convert the mass of hydrogen from grams to moles by using the equation moles = mass / molar mass
Now that we have the moles of hydrogen, we can use the molar ratio given in the equation of 1:3. Since we need 1/3 of the moles of hydrogen as nitrogen, we know we need 3.33 moles of nitrogen gas.
Now, we simply multiply the moles by the molar mass of nitrogen gas (28 grams per mol) and we find we need 93.33 grams of nitrogen gas.
Example Question #2 : Stoichiometry With Reactions
The formation of ammonia is given by the following equation:
Assuming you start with 15mol of hydrogen gas, how many moles of nitrogen gas will be needed to completely react the hydrogen gas?
Since the amount of hydrogen gas is given in moles, this solution is as easy as using the molar ratio between the reactants. Looking at the equation, we see that there is a 1:3 ratio for nitrogen to hydrogen. As a result, we would need one third of the amount of hydrogen available in order to use up all of the hydrogen gas.
Example Question #4 : Stoichiometry With Reactions
How many moles of water vapor will be formed when 5.00g of hydrogen gas react with excess oxygen?
When dealing with questions related to stoichiometry, it is important to compare only mole-to-mole ratios.
Hydrogen gas is diatomic. When converting grams to moles, we will need to double the atomic mass of hydrogen.
We now know we have 2.5 moles of hydrogen gas. Now we need to write a balanced equation to determine the conversion factor between hydrogen gas and the product.
For every two moles of hydrogen consumed, two moles of water are produced. Using this ratio and the starting moles of hydrogen, we can calculate the moles of water produced.