When 0.100 mol of carbon is burned in a close vessel with 8.00g of oxygen, how many grams of carbon dioxide can form? Which reactant is in excess, and how many grams of it remain after the reaction?

C + O2 ==> CO2

You have 0.1 mol C.
Convert 8 g O2 to moles. 8/32 = 0.25 mol.

0.1 mole C can form 0.1 mole CO2
0.25 mole O2 can form 0.25 mole CO2.
In limiting reagent problems, the smaller number is ALWAYS the correct one to choose. So C is the limiting reagent and will react completely. Some oxygen will remain unreacted.

g CO2 formed = 0.1 mole CO2 x molar mass CO2 = 0.1 x 44 = ?? grams CO2.

How much oxygen remains. First, determine how much oxygen is used.
0.1 mols CO2 will use 0.1 mole O2 and 0.1 mole O2 is 0.1 x 32 = 3.2 grams.
We had 8 g initially; therefore, the amount remaining is 8.0 - 3.2 = ?? grams. Check my work carefully. It's getting late here.

Well, it seems we have a case of carbon and oxygen getting up to some combustion shenanigans! Let's get calculating.

First, we need to convert the given amount of carbon (0.100 mol) into grams. The molar mass of carbon is about 12.01 g/mol, so 0.100 mol of carbon is equal to approximately 1.201 grams.

Now, let's take a look at the reaction equation for carbon combustion:

C + O2 → CO2

For every 1 mole of carbon burned, 1 mole of carbon dioxide is formed. So, the number of moles of carbon dioxide formed will also be 0.100 mol.

The molar mass of carbon dioxide (CO2) is approximately 44.01 g/mol. Therefore, 0.100 mol of carbon dioxide would be equal to about 4.401 grams.

Now, let's move on to the amount of oxygen present. We were given that there are 8.00 grams of oxygen. The molar mass of oxygen is approximately 16.00 g/mol.

To find the number of moles of oxygen, we need to divide the mass (8.00 g) by the molar mass (16.00 g/mol). This gives us 0.500 moles of oxygen.

According to the reaction equation, the ratio of carbon to oxygen is 1:O2. Since we have 0.100 moles of carbon and 0.500 moles of oxygen, it seems we have an excess of oxygen.

To determine how much excess oxygen remains after the reaction, we need to find out how much oxygen was actually used. The ratio of carbon to oxygen in the equation is 1:O2, so for every mole of carbon burned, we need 1 mole of oxygen.

Since we have 0.100 moles of carbon, we would need 0.100 moles of oxygen. However, we have 0.500 moles of oxygen, which means we have an excess of 0.500 - 0.100 = 0.400 moles of oxygen.

Now, to convert this excess amount of oxygen into grams, we can multiply it by the molar mass (16.00 g/mol). So, 0.400 moles of oxygen would be equal to approximately 6.40 grams.

To summarize:
- The grams of carbon dioxide formed is approximately 4.401 g.
- The excess reactant is oxygen.
- The grams of excess oxygen remaining after the reaction is approximately 6.40 g.

Phew, quite the circus act with carbon and oxygen! I hope this helps! Let me know if you have any more questions (or jokes)!

To determine how many grams of carbon dioxide can form, we first need to write and balance the chemical equation for the combustion of carbon:

C + O2 -> CO2

The balanced equation tells us that one mol of carbon reacts with one mol of oxygen to yield one mol of carbon dioxide.

From the equation, we can see that the molar ratio between carbon and carbon dioxide is 1:1. This means that if 0.100 mol of carbon is burned, we would expect to form 0.100 mol of carbon dioxide.

To determine the mass of carbon dioxide formed, we need to use the molar mass of carbon dioxide, which is approximately 44.01 g/mol.

Mass of carbon dioxide = moles of carbon dioxide * molar mass of carbon dioxide
= 0.100 mol * 44.01 g/mol
= 4.401 g

Therefore, 4.401 grams of carbon dioxide can form.

To determine which reactant is in excess, we need to compare the amount of oxygen consumed in the reaction with the amount of oxygen available.

The molar mass of oxygen is approximately 16.00 g/mol.

Mass of oxygen = 8.00 g

Moles of oxygen = mass of oxygen / molar mass of oxygen
= 8.00 g / 16.00 g/mol
= 0.500 mol

From the balanced equation, we know that one mol of carbon reacts with one mol of oxygen. Therefore, we need 0.100 mol of oxygen to react completely with 0.100 mol of carbon.

Since we have 0.500 mol of oxygen available, we can see that there is excess oxygen.

To find out how many grams of excess oxygen remain after the reaction, we need to find the difference between the initial moles of oxygen and the moles of oxygen consumed in the reaction.

Moles of excess oxygen = moles of oxygen available - moles of oxygen consumed
= 0.500 mol - 0.100 mol
= 0.400 mol

Mass of excess oxygen = moles of excess oxygen * molar mass of oxygen
= 0.400 mol * 16.00 g/mol
= 6.4 g

Therefore, there are 6.4 grams of excess oxygen that remain after the reaction.

To determine the number of grams of carbon dioxide that can form, we need to first calculate the limiting reactant. The limiting reactant is the one that is completely consumed and determines the amount of product formed.

To do this, we need to compare the number of moles of carbon and oxygen to determine which one is limiting.

1. Calculate the moles of carbon:
Moles of carbon = mass of carbon (in grams) / molar mass of carbon

The molar mass of carbon (C) is 12.01 g/mol. Since we have 0.100 mol of carbon, the mass of carbon is given as:
Mass of carbon = 0.100 mol × 12.01 g/mol

2. Calculate the moles of oxygen:
Moles of oxygen = mass of oxygen (in grams) / molar mass of oxygen

The molar mass of oxygen (O₂) is 32.00 g/mol. We are given that there are 8.00 g of oxygen, so the moles of oxygen can be calculated as:
Moles of oxygen = 8.00 g / 32.00 g/mol

3. Calculate the reaction ratio:
The balanced equation for the reaction between carbon and oxygen is:
C + O₂ → CO₂

From the balanced equation, we can see that the mole ratio between carbon and oxygen is 1:1. Therefore, if the number of moles of carbon and oxygen is equal, neither reactant is in excess.

4. Determine the limiting reactant:
Compare the moles of carbon and oxygen calculated in steps 1 and 2. The reactant with the lower number of moles is the limiting reactant.

Now that we obtained the moles of carbon and oxygen, we can compare them:
Moles of carbon = 0.100 mol
Moles of oxygen = (8.00 g / 32.00 g/mol) = 0.250 mol

Since carbon has the lower number of moles (0.100 mol < 0.250 mol), it is the limiting reactant.

5. Calculate the moles of carbon dioxide formed:
From the balanced equation, we know that 1 mole of carbon reacts to form 1 mole of carbon dioxide.

Therefore, the number of moles of carbon dioxide formed is equal to the number of moles of carbon:
Moles of carbon dioxide = Moles of carbon = 0.100 mol

6. Calculate the mass of carbon dioxide formed:
To calculate the mass of carbon dioxide formed, we need to multiply the number of moles by the molar mass of carbon dioxide.

The molar mass of carbon dioxide (CO₂) is 44.01 g/mol.

Mass of carbon dioxide = Moles of carbon dioxide × molar mass of carbon dioxide
= 0.100 mol × 44.01 g/mol

The calculated mass of carbon dioxide will give us the grams of carbon dioxide formed during the reaction.

To determine which reactant is in excess, we compare the amount of oxygen actually used in the reaction to the initial amount of oxygen given.

7. Calculate the remaining excess reactant:
To find the remaining excess reactant, we need to subtract the moles of the limiting reactant from the moles of the excess reactant.

Moles of remaining excess reactant = Initial moles of excess reactant - Moles of limiting reactant

The initial moles of excess reactant (oxygen) is the moles of oxygen calculated in step 2:
Initial moles of oxygen = 0.250 mol

Therefore,
Moles of remaining excess reactant = 0.250 mol - 0.100 mol

To calculate the mass of the remaining excess reactant, we multiply the moles by the molar mass of oxygen.

8. Calculate the mass of the remaining excess reactant:
Mass of remaining excess reactant = Moles of remaining excess reactant × molar mass of oxygen

Now, you can perform the calculations mentioned above to find the grams of carbon dioxide formed and the mass of the remaining excess reactant (oxygen) after the reaction.