Consider the following balanced reaction. What mass (in g) of CO2 can be formed from 743 mg of O2? Assume that there is excess C3H7SH present.

C3H7SH (l) + 6 O2 (g) ---> 3 CO2 (g) + SO2 (g) + 4 H20 (g)

Convert 743 mg O2 to grams, then to moles. moles = gram/molar mass

Using the coefficients in the balanced equation, convert moles O2 to moles CO2,

Now convert moles CO2 to grams CO2. g = moles x molar mass.

To solve this problem, we need to determine the stoichiometry of the reaction and use it to calculate the mass of CO2 formed.

First, let's convert the given mass of O2 from milligrams (mg) to grams (g). Since 1 gram (g) is equal to 1000 milligrams (mg), we have:

743 mg = 743/1000 = 0.743 g

Next, we need to use the balanced equation to find the molar ratio between O2 and CO2. According to the balanced equation, 6 moles of O2 react to form 3 moles of CO2. This means that for every 6 moles of O2, 3 moles of CO2 are produced.

Now, we need to convert the mass of O2 to moles using its molar mass. The molar mass of O2 is 32 g/mol (16 g/mol for each oxygen atom). Therefore, the number of moles of O2 can be calculated as:

moles of O2 = mass of O2 / molar mass of O2 = 0.743 g / 32 g/mol = 0.023 moles of O2

Now, we can use the stoichiometry to determine the moles of CO2 formed. For every 6 moles of O2, 3 moles of CO2 are produced. Therefore, we can set up a ratio:

moles of CO2 / moles of O2 = 3/6 = 1/2

moles of CO2 = moles of O2 × (1/2) = 0.023 moles of O2 × (1/2) = 0.0115 moles of CO2

Finally, we can convert the moles of CO2 to grams by multiplying with its molar mass. The molar mass of CO2 is 44 g/mol (12 g/mol for each carbon atom and 16 g/mol for each oxygen atom).

mass of CO2 = moles of CO2 × molar mass of CO2 = 0.0115 moles of CO2 × 44 g/mol = 0.506 g

Therefore, the mass of CO2 that can be formed from 743 mg of O2 is approximately 0.506 g.