When the sugar glucose, C6H12O6 is burned in air, carbon dioxide and water vapor are produced. Write the balanced chemical equation for this process, and calculate the theoretical yield of carbon dioxide when 1.00 g of glucose is burned completely.

google is your friend. First hit was

C6H12O6 + 6O2 ==> 6CO2 + 6H2O

now work with moles, then convert back to grams.

Idk man seems hard

c6h1206 + 6o2 -> 6co2 + 6h2o

1.47g

How do you convert this problem into grams?

To write the balanced chemical equation for the combustion of glucose, we need to know the stoichiometric ratio of the reactants and products involved in the reaction. The combustion of glucose can be represented by the following balanced equation:

C6H12O6 + 6O2 → 6CO2 + 6H2O

In this equation, one molecule of glucose (C6H12O6) reacts with six molecules of oxygen (O2) to produce six molecules of carbon dioxide (CO2) and six molecules of water (H2O).

To calculate the theoretical yield of carbon dioxide when 1.00 g of glucose is burned completely, we need to use the molar mass of glucose and convert it to moles.

The molar mass of glucose (C6H12O6) can be calculated as follows:
(6 x molar mass of carbon) + (12 x molar mass of hydrogen) + (6 x molar mass of oxygen)

= (6 x 12.01 g/mol) + (12 x 1.01 g/mol) + (6 x 16.00 g/mol)
= 72.06 g/mol + 12.12 g/mol + 96.00 g/mol
= 180.18 g/mol

Now, we can calculate the number of moles of glucose in 1.00 g using the formula:
moles = mass / molar mass

moles = 1.00 g / 180.18 g/mol
moles ≈ 0.005550 mol (rounded to five decimal places)

From the balanced equation, we know that for every one mole of glucose, six moles of carbon dioxide are produced. Therefore, we can calculate the number of moles of carbon dioxide produced:

moles of CO2 = 6 x moles of glucose
moles of CO2 = 6 x 0.005550 mol
moles of CO2 ≈ 0.0333 mol (rounded to four decimal places)

Finally, to convert moles of carbon dioxide to grams, we can multiply by the molar mass of carbon dioxide, which is approximately 44.01 g/mol:

mass of CO2 = moles of CO2 x molar mass of CO2
mass of CO2 = 0.0333 mol x 44.01 g/mol
mass of CO2 ≈ 1.46 g (rounded to two decimal places)

Thus, the theoretical yield of carbon dioxide when 1.00 g of glucose is burned completely is approximately 1.46 grams.