If 7.50 grams 1-butene reacts with 25.0 g oxygen how many grams of carbon dioxide will form? Which reactant is limiting? How many grams of excess reactant will be left at the end of the reaction?

How do you know this is a limiting reagent problem. Because BOTH reactants are given, not just one of them.

1. Write the equation and balance it.
2. Convert 7.50 g 1-butene to moles. moles = grams/molar mass.
3. Convert 25.0 g oxygen to moles by the same procedure.
4. Using the coefficients in the balanced equation, convert moles butene to moles either product.
5. Same procedure, convert g oxygen to grams of the (same) product.
6. It is likely that the answer to 4 and 5 will not agree which means one is wrong; the correct answer in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent.
7. Now that you have identified the limiting reagent, that means the other reactant is the one that will have an excess. Using the coefficients in the balanced equation, convert moles of the limiting reagent to moles of the "other" reagent, convert that to grams (g = moles x molar mass) and subtract from the initial to find the excess (the amount that did not react).

To determine the grams of carbon dioxide formed, we need to balance the given chemical equation and use stoichiometry.

The balanced equation for the combustion of 1-butene (C₄H₈) is as follows:

C₄H₈ + 6O₂ → 4CO₂ + 4H₂O

From the equation, we can see that 1 mole of 1-butene reacts with 6 moles of oxygen to produce 4 moles of carbon dioxide.

Step 1: Convert the given mass of 1-butene to moles.
Given mass of 1-butene = 7.50 grams
Molar mass of 1-butene (C₄H₈) = 56.11 g/mol
Number of moles of 1-butene = 7.50 g / 56.11 g/mol

Step 2: Convert the given mass of oxygen to moles.
Given mass of oxygen = 25.0 grams
Molar mass of oxygen (O₂) = 32.00 g/mol
Number of moles of oxygen = 25.0 g / 32.00 g/mol

Step 3: Determine the limiting reactant.
To determine the limiting reactant, we compare the mole ratios based on the balanced equation. The reactant that produces the lesser amount of product is the limiting reactant.

From the balanced equation, we can see that 1 mole of 1-butene reacts with 6 moles of oxygen, which means that:
Moles of 1-butene : Moles of oxygen = 1 : 6

We have calculated the number of moles for both reactants. Now we can compare them to determine the limiting reactant.

Step 4: Calculate the grams of carbon dioxide formed.
Using stoichiometry, we can calculate the grams of carbon dioxide produced based on the limiting reactant. Since the limiting reactant determines the amount of product formed, we use its moles to calculate the grams of carbon dioxide.

Step 5: Determine the grams of excess reactant left.
After determining the limiting reactant and calculating the grams of carbon dioxide produced, we can calculate the grams of excess reactant left. This can be done by subtracting the amount of the limiting reactant used from the amount initially given for the excess reactant.

To calculate these values, we need to know the number of moles of oxygen and 1-butene. Please provide the molar masses of both compounds so we can continue the calculations.