What is theoretical yield in grams of CO2 formed from reaction of 3.12g CS2 and 1.88g O2?

convert each to moles.

Write the balanced equation.
The reagent with the smaller moles used will limit the reaction.

So, figure the moles produced, and convert back to grams.

THe answer will be = 2.585 gram

To find the theoretical yield of CO2 formed in grams, you need to follow these steps:

Step 1: Write and balance the chemical equation for the reaction.
CS2 + 3O2 -> CO2 + 2SO2

Step 2: Calculate the molar mass of each compound.
Molar mass of CS2 = 12.01 g/mol (C) + 32.07 g/mol (S) x 2 = 76.14 g/mol
Molar mass of O2 = 16.00 g/mol (O) x 2 = 32.00 g/mol
Molar mass of CO2 = 12.01 g/mol (C) + 16.00 g/mol (O) x 2 = 44.01 g/mol

Step 3: Convert the given masses into moles.
moles of CS2 = 3.12 g / 76.14 g/mol = 0.0409 mol
moles of O2 = 1.88 g / 32.00 g/mol = 0.0588 mol

Step 4: Determine the limiting reagent.
To find the limiting reagent, compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The reactant that produces the least moles of the product is the limiting reagent.

From the balanced equation, 1 mole of CS2 reacts with 3 moles of O2 to produce 1 mole of CO2. Therefore, the stoichiometric ratio of CS2 to O2 is 1:3.

moles of O2 needed = 0.0409 mol CS2 x (3 mol O2 / 1 mol CS2) = 0.1227 mol

Since we have 0.0588 mol of O2 available, O2 is the limiting reagent.

Step 5: Calculate the theoretical yield of CO2.
Theoretical yield of CO2 = moles of O2 x molar mass of CO2
Theoretical yield of CO2 = 0.0588 mol x 44.01 g/mol = 2.569 g

Therefore, the theoretical yield of CO2 formed from the reaction of 3.12 g CS2 and 1.88 g O2 is 2.569 grams.

To find the theoretical yield of CO2 formed from the reaction between CS2 and O2, we need to follow these steps:

Step 1: Write the balanced chemical equation for the reaction.
In this case, the reaction can be represented by the following equation:

CS2 + 3O2 -> CO2 + 2SO2

Step 2: Calculate the molar mass of each compound.
CS2: 1 carbon (C) atom with a molar mass of 12.01 g/mol + 2 sulfur (S) atoms with a molar mass of 32.06 g/mol = 76.14 g/mol
O2: 2 oxygen (O) atoms with a molar mass of 16.00 g/mol = 32.00 g/mol
CO2: 1 carbon (C) atom with a molar mass of 12.01 g/mol + 2 oxygen (O) atoms with a molar mass of 16.00 g/mol = 44.01 g/mol

Step 3: Calculate the number of moles of each reactant.
Number of moles = mass / molar mass

For CS2:
Number of moles of CS2 = 3.12 g / 76.14 g/mol

For O2:
Number of moles of O2 = 1.88 g / 32.00 g/mol

Step 4: Identify the limiting reactant.
To determine the limiting reactant, we compare the stoichiometric ratios between the reactants. From the balanced equation, we can see that the stoichiometric ratio between CS2 and O2 is 1:3. This means that for every 1 mole of CS2, we need 3 moles of O2 for the reaction to go to completion.

Calculate the moles of CO2 that can be formed from each reactant by multiplying the number of moles of each reactant by the stoichiometric ratio:

For CS2:
Moles of CO2 from CS2 = (3.12 g / 76.14 g/mol) × (1 mol CO2 / 1 mol CS2)

For O2:
Moles of CO2 from O2 = (1.88 g / 32.00 g/mol) × (1 mol CO2 / 3 mol O2)

The limiting reactant is the one that produces the fewer moles of CO2. So, we compare the moles of CO2 from both reactants and choose the smaller value.

Step 5: Calculate the theoretical yield of CO2.
Theoretical yield of CO2 = moles of CO2 (from limiting reactant) × molar mass of CO2

Now you can substitute the calculated values into the equation and solve for the theoretical yield of CO2 in grams.