Other Stoichiometric Calculations - AP Chemistry

Each mole of methane to react releases 5kJ of heat. Since we have 18 moles of methane, we expect a potential of 90kJ of heat to be released.

We also have to take into account that the process is only 60% efficient; therefore, we effectively only liberate 60% of the total expected amount.

The net ionic equation that occurs is:

and are spectator ions. Hence, concentration is not affected. The molecular mass of is , then of NaCl yields of . Since we have of a solution of , we have of . There is no limiting reactant. Hence of are formed. Being the molecular mass of , are formed. Result should be given with tree significant figures.

Remember: the given measurement has 4 significant figures, so the answer must also have 4 significant figures.

Remember, there really isn't a way to go straight from grams to atoms, but, it's possible to change grams into moles (using the molar mass), and it's possible to change moles into atoms (using Avogadro's number).

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 :

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 :

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 20.0 grams, the moles of 20.0 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, for every mole of the compound , there is 1 mole of bromine. Based on our calculations 20.0 grams of is 0.1835 moles of the compound. Because there is a 1:1 molar ratio of bromine to every molecule, we have 0.1835 moles of bromine in 20 grams of our compound, .

First we must determine the volume of the container:

The density is .

Now we will convert this into grams per milliliter.

Each mole of methane to react releases 5kJ of heat. Since we have 18 moles of methane, we expect a potential of 90kJ of heat to be released.

We also have to take into account that the process is only 60% efficient; therefore, we effectively only liberate 60% of the total expected amount.

The net ionic equation that occurs is:

and are spectator ions. Hence, concentration is not affected. The molecular mass of is , then of NaCl yields of . Since we have of a solution of , we have of . There is no limiting reactant. Hence of are formed. Being the molecular mass of , are formed. Result should be given with tree significant figures.

Remember: the given measurement has 4 significant figures, so the answer must also have 4 significant figures.

Remember, there really isn't a way to go straight from grams to atoms, but, it's possible to change grams into moles (using the molar mass), and it's possible to change moles into atoms (using Avogadro's number).

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 :

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 :

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 20.0 grams, the moles of 20.0 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, for every mole of the compound , there is 1 mole of bromine. Based on our calculations 20.0 grams of is 0.1835 moles of the compound. Because there is a 1:1 molar ratio of bromine to every molecule, we have 0.1835 moles of bromine in 20 grams of our compound, .

First we must determine the volume of the container:

The density is .

Now we will convert this into grams per milliliter.

Each mole of methane to react releases 5kJ of heat. Since we have 18 moles of methane, we expect a potential of 90kJ of heat to be released.

We also have to take into account that the process is only 60% efficient; therefore, we effectively only liberate 60% of the total expected amount.

The net ionic equation that occurs is:

and are spectator ions. Hence, concentration is not affected. The molecular mass of is , then of NaCl yields of . Since we have of a solution of , we have of . There is no limiting reactant. Hence of are formed. Being the molecular mass of , are formed. Result should be given with tree significant figures.

Remember: the given measurement has 4 significant figures, so the answer must also have 4 significant figures.