Stoichiometry Problem: A.) The mitochondria in our cells, through the process of

Question

Stoichiometry Problem: A.) The mitochondria in our cells, through the process of cellular respiration, convert glucose to the high energy molecule ATP. This combustion reaction also produces the benign waste products water and carbon dioxide. If only 3,000 grams of oxygen are available, how much glucose would combust?

Step 1- Balance equation- (Already done for you!) C6H12O6 + 6O2 ATP + 6CO2 + 6H2O

Step 2- Grams-to-Moles of Known-

Step 3- Ratio-for moles unknown

Step 4- Moles-to-Grams of unknown

B.) Stoichiometry II: This is volume-volume, so it should be one step! Nitrogen monoxide (NO) combines with oxygen to form nitrogen dioxide (NO2). 2 NO(g) + O2(g) 2NO2(g)

If 15 liters of oxygen are used, what is the volume of nitrogen dioxide that will be produced?

Explanation / Answer

Step 1- Balance equation

C6H12O6 + 6O2 + ATP ------> 6CO2 + 6H2O

Step 2- Grams-to-Moles of Known

no. of moles = weight (gms)/mol.wt(g/mol) = 3000/32 = 93.75 mole of O2

Step 3- Ratio-for moles unknown

Form the stoichiometry of equation it is clear that 6 moles of O2 requires 1mole of Glucose . Thus, 93.75mole of O2 requires 93.75mol*1mol/6mol = 15.625 mole of C6H12O6

Step 4- Moles-to-Grams of unknown

The amount of C6H12O6 required is given by

amount(gms) = no.of moles*mol.wt(g/mol) = 15.625*180 = 2812.5gms

2) 2 NO(g) + O2(g) ----> 2NO2(g)

form stoichiometry of equation 2volumes of NO requires 1 volume of O2 to form 2 volumes of NO2

If 15L of O2 are used, 2*15= 30L of NO2 will be produced

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