What mass of carbon monoxide will be produce if 40 grams of carbon are all allowed to react with 95 grams of sulfur dioxide?
To determine the mass of carbon monoxide produced in the reaction, we need to first write and balance the chemical equation for the reaction between carbon and sulfur dioxide.
The balanced chemical equation for the reaction is as follows:
C + 2SO2 -> COSO2
From the balanced equation, we can see that for every one mole of carbon, we react with two moles of sulfur dioxide to produce one mole of carbon monoxide.
Now, let's calculate the number of moles of carbon and sulfur dioxide present in the given masses:
Mass of Carbon (C) = 40 grams
Molar Mass of Carbon (C) = 12.01 g/mol
Number of Moles of Carbon (nC) = Mass of Carbon / Molar Mass of Carbon
= 40 g / 12.01 g/mol
≈ 3.33 mol (rounded to two decimal places)
Mass of Sulfur Dioxide (SO2) = 95 grams
Molar Mass of Sulfur Dioxide (SO2) = 32.06 g/mol + (2 * 16.00 g/mol)
= 64.06 g/mol
Number of Moles of Sulfur Dioxide (nSO2) = Mass of Sulfur Dioxide / Molar Mass of Sulfur Dioxide
= 95 g / 64.06 g/mol
≈ 1.48 mol (rounded to two decimal places)
Now, we can determine the limiting reactant, which is the reactant that is completely consumed in the reaction and determines the amount of product formed.
From the balanced equation, the stoichiometric ratio between carbon and sulfur dioxide is 1:2. Therefore, for complete reaction, we require twice as many moles of sulfur dioxide as carbon.
Since we have 3.33 moles of carbon and 1.48 moles of sulfur dioxide, it is clear that sulfur dioxide is limiting. Hence, carbon is in excess.
To calculate the mass of carbon monoxide produced, we need to know the number of moles of carbon monoxide formed, which can be determined from the limiting reactant (in this case, sulfur dioxide).
According to the balanced equation, the stoichiometric ratio between sulfur dioxide and carbon monoxide is 2:1. Thus, for every two moles of sulfur dioxide, we produce one mole of carbon monoxide.
Number of Moles of Carbon Monoxide (nCO) = (1/2) * Number of Moles of Sulfur Dioxide
= (1/2) * 1.48 mol
= 0.74 mol
To calculate the mass of carbon monoxide produced, we can use the molar mass of carbon monoxide.
Molar Mass of Carbon Monoxide (CO) = 12.01 g/mol + 16.00 g/mol
= 28.01 g/mol
Mass of Carbon Monoxide (CO) = Number of Moles of Carbon Monoxide * Molar Mass of Carbon Monoxide
= 0.74 mol * 28.01 g/mol
≈ 20.68 grams (rounded to two decimal places)
Therefore, approximately 20.68 grams of carbon monoxide will be produced in the reaction.
To determine the mass of carbon monoxide (CO) produced, we need to first write the balanced chemical equation for the reaction between carbon (C) and sulfur dioxide (SO₂):
C + SO₂ → CO
The molar mass of carbon is 12 g/mol and the molar mass of sulfur dioxide is 32 g/mol + 16 g/mol × 2 = 64 g/mol.
To calculate the amount of carbon monoxide produced, we can use the concept of stoichiometry and the given masses of carbon and sulfur dioxide.
1. Calculate the number of moles of carbon:
Moles of carbon = mass of carbon / molar mass of carbon
= 40 g / 12 g/mol
≈ 3.33 mol
2. Calculate the number of moles of sulfur dioxide:
Moles of sulfur dioxide = mass of sulfur dioxide / molar mass of sulfur dioxide
= 95 g / 64 g/mol
≈ 1.48 mol
3. Determine the limiting reactant:
The reactant that is used up first will determine the maximum amount of product that can be formed. To determine the limiting reactant, we compare the number of moles of carbon and sulfur dioxide. In this case, sulfur dioxide is the limiting reactant since it has the smallest number of moles.
4. Use the stoichiometry of the balanced equation to find the number of moles of carbon monoxide produced:
From the balanced equation, we see that one mole of carbon reacts with one mole of sulfur dioxide to produce one mole of carbon monoxide.
Therefore, the number of moles of carbon monoxide produced is equal to the number of moles of sulfur dioxide, which is approximately 1.48 mol.
5. Calculate the mass of carbon monoxide produced:
Mass of carbon monoxide = number of moles of carbon monoxide × molar mass of carbon monoxide
= 1.48 mol × (12 g/mol + 16 g/mol)
≈ 43.2 g
Therefore, approximately 43.2 grams of carbon monoxide will be produced when 40 grams of carbon react with 95 grams of sulfur dioxide.