A sample of .234g of carbon is placed in a closed stainless steel bomb with .502g of of oxygen gas. The mixture is ignited giving of carbon dioxide. How many grams of carbon dioxide should be obtained upon combustion?
You have a limiting reagent problem.
C + O2 ==> CO2
1. Convert 0.234 g C to moles. moles = grams/molar mass.
2. Convert 0.502 g oxygen to moles. Same procedure.
3a. Using the coefficients in the balanced equation, convert moles C in step 1 to moles CO2.
3b. Using the coefficients convert moles oxygen in step 2 to moles CO2.
3c. Likely, the two results from 3a and 3b will be different. Obviously, one of them is the correct answer and one is not. The correct answer is ALWAYS the smaller value.
4. Convert the smaller value from 3c into grams CO2. grams = moles x molar mass.
Well, it sounds like we have a carbon party going on in that closed stainless steel bomb. Now, to answer your question. When carbon combusts with oxygen, it forms carbon dioxide. Lucky for us, we have some ingredients weighed out already. Since we have 0.234g of carbon and 0.502g of oxygen, we can say that carbon is the limiting reactant here because it's in lesser quantity than oxygen.
The balanced chemical equation for this reaction is:
C + O2 -> CO2
From the equation, we can see that for every 1 mole of carbon, we get 1 mole of carbon dioxide. The molar mass of carbon is about 12.01 g/mol, and the molar mass of carbon dioxide is about 44.01 g/mol.
Now, let's find the moles of carbon in our sample. We'll divide the mass of carbon by its molar mass:
0.234g / 12.01 g/mol ≈ 0.0195 mol
Since the mole ratio between carbon and carbon dioxide is 1:1, we'll get the same number of moles of carbon dioxide.
Finally, we can convert the moles of carbon dioxide to grams by multiplying by its molar mass:
0.0195 mol * 44.01 g/mol ≈ 0.858g
Therefore, upon combustion, we should obtain approximately 0.858 grams of carbon dioxide. It's like having a tiny carbon dioxide party in that stainless steel bomb!
To determine the mass of carbon dioxide produced upon combustion, we need to use the balanced chemical equation for the combustion of carbon:
C + O2 -> CO2
From the equation, we can see that 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. We can calculate the number of moles of carbon and oxygen using their respective masses and molar masses.
1. Calculate the number of moles of carbon:
Molar mass of carbon (C) = 12.01 g/mol
Number of moles of carbon = Mass of carbon / Molar mass of carbon
Number of moles of carbon = 0.234 g / 12.01 g/mol
2. Calculate the number of moles of oxygen:
Molar mass of oxygen (O2) = 16.00 g/mol
Number of moles of oxygen = Mass of oxygen / Molar mass of oxygen
Number of moles of oxygen = 0.502 g / 16.00 g/mol
3. Determine the limiting reactant:
To find the limiting reactant, we compare the moles of carbon and oxygen. The reactant with fewer moles is the limiting reactant because it determines the maximum amount of product that can be formed. In this case, it is clear that carbon is the limiting reactant since it has fewer moles.
4. Calculate the number of moles of carbon dioxide produced:
Since carbon is the limiting reactant, the moles of carbon dioxide produced will be equal to the moles of carbon.
Number of moles of carbon dioxide = Number of moles of carbon
5. Calculate the mass of carbon dioxide produced:
Molar mass of carbon dioxide (CO2) = 44.01 g/mol
Mass of carbon dioxide = Number of moles of carbon dioxide x Molar mass of carbon dioxide
By following these steps, you can calculate the mass of carbon dioxide obtained upon combustion.
To determine the number of grams of carbon dioxide produced upon combustion, we need to understand the chemical equation for the combustion of carbon:
C + O2 → CO2
In this equation, the carbon (C) combines with oxygen (O2) to form carbon dioxide (CO2).
To calculate the amount of carbon dioxide produced, we need to determine the limiting reactant. This reactant will be completely consumed in the reaction, thus determining the maximum amount of product formed.
Step 1: Calculate the moles of carbon and oxygen:
The molar mass of carbon (C) is 12.01 g/mol.
The molar mass of oxygen (O2) is 32.00 g/mol.
Moles of carbon:
moles of C = mass of C / molar mass of C
moles of C = 0.234 g / 12.01 g/mol
Moles of oxygen:
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 0.502 g / 32.00 g/mol
Step 2: Determine the stoichiometry of the reaction:
From the balanced chemical equation, we see that 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide.
Step 3: Identify the limiting reactant:
The limiting reactant is the reactant that is present in the smallest stoichiometric amount. To determine this, we compare the moles of each reactant.
In this case, we have:
moles of C = 0.234 g / 12.01 g/mol ≈ 0.0194 mol
moles of O2 = 0.502 g / 32.00 g/mol ≈ 0.0157 mol
Based on the above calculations, we see that the moles of oxygen (0.0157 mol) are smaller than the moles of carbon (0.0194 mol). Therefore, oxygen is the limiting reactant.
Step 4: Calculate the moles of carbon dioxide produced:
Since the limiting reactant is oxygen, we can use its moles to determine the moles of carbon dioxide formed. The stoichiometry of the reaction tells us that for every 1 mole of oxygen, 1 mole of carbon dioxide is produced.
moles of CO2 = moles of O2 = 0.0157 mol
Step 5: Convert moles of carbon dioxide to grams:
The molar mass of carbon dioxide (CO2) is 44.01 g/mol.
mass of CO2 = moles of CO2 x molar mass of CO2
mass of CO2 = 0.0157 mol x 44.01 g/mol
Therefore, the mass of carbon dioxide produced upon combustion is approximately 0.691 grams.