The molar mass of glucose is 180.2 g/mol. How many grams of glucose will be produced when 132.0 g of CO2 reacts with an excess of water?

The mass in grams of glucose produced when 132.0 g of CO2 reacts with an excess of water is 90.1 grams

Explanation:

The chemical equation for the reaction is

6H₂O + 6CO₂  → C₆H₁₂O₆ + 6O₂

From the reaction, it is seen that 6 moles of H₂O reacts ith 6 moles of CO₂ to produce 1 mole of glucose  C₆H₁₂O₆ and 6 moles oxygen gas

The molar mass of CO₂ = 44.01 g/mol

There fpre 132.0 g contains 132.0/44.01 moles or ≅ 3 moles

However since 6 moles of CO₂ produces 1 mole of O₂, then 3 moles of CO₂ will prduce 1/6×3 or 0.5 moles of C₆H₁₂O₆

and since the molar mass (or the mass of one mole) of C₆H₁₂O₆ is 180.2 grams/mole then 0.5 mole of C₆H₁₂O₆ will have a mass of

mass of 1 mole C₆H₁₂O₆ = 180.2 g

mass of 0.5 mole C₆H₁₂O₆ = 180.2 g × 0.5 = 90.1 grams

Mass of glucose produced = 90.1 grams

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.