Consider the complete combustion of glucose (c6h12o6) with o2 and calculate the moles of co2 produced when 1.02 g of glucose is reacted with 25 ml of o2 at body temperature (37 ºc) and 0.970 atm.

The moles of carbon dioxide produced is 0.00095 moles

Explanation:

To calculate the number of moles, we use the equation:

Given mass of glucose = 1.02 g

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

To calculate the number of moles, we use ideal gas equation:

where,

P = pressure of the gas = 0.970 atm  

V = Volume of the gas = 25 mL = 0.025 L   (Conversion factor:  1 L = 1000 mL)

T = Temperature of the gas =

R = Gas constant =

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

The chemical equation for the combustion of glucose follows:

By Stoichiometry of the reaction:

6 moles of oxygen gas reacts with 1 mole of glucose

So, 0.00095 moles of oxygen gas will react with = of glucose

As, given amount of glucose is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

6 moles of oxygen gas produces 6 moles of carbon dioxide

So, 0.00095 moles of oxygen gas will produce = of carbon dioxide

Hence, the moles of carbon dioxide produced is 0.00095 moles

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)

6.4 cm (max) - 25 cm (min)