When cane sugar reacts with oxygen in living systems, carbon dioxide and water are produced. What weight of carbon dioxide can be produced from the reaction of 15.0 grams of cane sugar with 15.0 grams of oxygen? (Hint: balance the equation)

___C12H22O11(s) + ___O2(g)  ___CO2(g) + ___H2O(g)

C12H22O11(s) + 12O2(g)==> 12CO2(g) + 11H2O(g)

You know this is another limiting reagent (LR) problem because amounts are given for BOTH reactants.
mols sugar = grams/molar mass = ?
mols O2 = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols sugar to mols CO2.
Do the same for converting mol O2 to mols CO2.
It is likely that the values you obtain will be different which means one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the regent producing the smaller value is called the LR.
Now use the smaller value to calculate g CO2. g CO2 = mol CO2 x molar mass CO2

You may want to go back and look at the post down the board for the 200 molecules H2 and 200 molecules O2. I corrected that answer.

thank you!

its 16.03

To find the weight of carbon dioxide produced from the reaction between cane sugar (C12H22O11) and oxygen (O2), we need to balance the chemical equation first.

The balanced chemical equation for the reaction is:

C12H22O11(s) + 12O2(g) -> 12CO2(g) + 11H2O(g)

By balancing the equation, we see that it takes 12 moles of O2 to react with 1 mole of C12H22O11, resulting in the production of 12 moles of CO2.

Now, let's calculate the molar masses of cane sugar (C12H22O11) and oxygen (O2):
- Molar mass of C12H22O11 = (12*12.01) + (22*1.01) + (11*16.00) = 342.34 g/mol
- Molar mass of O2 = (2*16.00) = 32.00 g/mol

Next, we need to determine the limiting reactant (the reactant that gets consumed completely and determines the amount of product formed). To do this, we compare the number of moles of each reactant to see which one is smaller.

Number of moles of C12H22O11 = Mass / Molar mass = 15.0 g / 342.34 g/mol ≈ 0.0438 mol
Number of moles of O2 = Mass / Molar mass = 15.0 g / 32.00 g/mol ≈ 0.469 mol

In this case, the moles of C12H22O11 is significantly smaller than the moles of O2. Therefore, C12H22O11 is the limiting reactant.

Now, we can use the balanced equation to find the moles of CO2 produced from the moles of C12H22O11. From the stoichiometry of the balanced equation, we know that 1 mole of C12H22O11 produces 12 moles of CO2.

Number of moles of CO2 = 0.0438 mol C12H22O11 × (12 mol CO2 / 1 mol C12H22O11) = 0.5256 mol CO2

Finally, to find the weight of CO2 produced, we calculate the mass of CO2 using the molar mass of CO2:

Mass of CO2 = Number of moles × Molar mass = 0.5256 mol × 44.01 g/mol ≈ 23.13 g

Therefore, approximately 23.13 grams of carbon dioxide can be produced from the reaction of 15.0 grams of cane sugar with 15.0 grams of oxygen.