Glucose, c6h12o6, is used as an energy source by the human body. the overall reaction in the body is described by the equation
c6h12o6 (aq) + 6o2 (g) = 6co2 (g) + 6h2o (l)
calculate the number of grams of oxygen required to convert 63.0 g of glucose to co2 and h2o.
o2
also compute the number of grams of co2 produced.
co2

We need 67.2 grams of 02. There  will be produced 92.421 grams of CO2

Explanation:

Step 1: The balanced equation

C6H12O6 (aq) + 6O2 (g) = 6CO2 (g) + 6H2O (l)

Step 2: Data given

We can see that for 1 mole C6H12O6 consumed, we need 6 moles of O2 to produce 6 moles of CO2 and 6 moles of H2O

The mass of glucose is 63.0g

The molar mass of glucose is 180.156 g/mole

The molar mass of O2 is 32 g/mole

The molar mass of CO2 is 44.01 g/mole

Step 3: Calculate number of moles of glucose

Number of moles of glucose = mass of glucose / molar mass of glucose

Number of moles of glucose = 63 grams /180.156 g/moles = 0.35 moles

Step 4: Calculate number of moles of oxygen ( and CO2)

For 1 mole of glucose we need 6 moles of O2 to produce 6 moles of CO2 and 6 moles of H2O

So for 0.35 moles of glucose we need 6*0.35 = 2.1 moles of O2 to produce 2.1 moles of CO2 and 2.1 moles of H2O

Step 5: Calculate mass of oxygen ( and CO2)

mass of oxygen = number of moles of oxygen * Molar mass of oxygen

mass of oxygen = 2.1 moles * 32 g/mole = 67.2 g O2 (≈ 67 grams)

mass of CO2 = number of moles of CO2 * Molar mass of CO2

mass of CO2 = 2.1 moles * 44.01 g/mole = 92.421 grams

We need 67.2 grams of 02. There  will be produced 92.421 grams of CO2

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.