Answer:
1,013 grams of oxygen gas will be consumed when 950g of glucose is oxidized completely.
Step-by-step explanation:
You have the balanced equation:
C₆H₁₂O₆ + 6 O₂ --> 6 H₂0 + 6 CO₂
First of all it is convenient to calculate the molar mass of the compounds participating in the reaction, knowing the atomic mass of each element:
- C: 12 g/mol
- H: 1 g/mol
- O: 16 g/mol
Then:
- C₆H₁₂O₆: 6*12 g/mol + 12*1 g/mol + 6*16 g/mol= 180 g/mol
- O₂: 2*16 g/mol= 32 g/mol
- H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol
- CO₂: 12 g/mol + 2*16 g/mol= 44 g/mol
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you can see that for 1 mole of glucose to react, 6 moles of oxygen are necessary.
Then, it is possible to determine the amounts of mass necessary for glucose and oxygen to react stoichiometrically:
- C₆H₁₂O₆: 1 mol*180 g/mol=180 g
- O₂: 6 moles*32 g/mol=192 g
Therefore, it is possible to apply a rule of three to calculate the amount of gaseous oxygen that will be consumed when 950 g of glucose is completely oxidized. The rule of three applies as follows: if by stoichiomatry 180 grams of glucose react with 192 grams of oxygen, when reacting 950 grams of glucose how many grams of oxygen will be needed?

grams of oxygen=1,013
So, 1,013 grams of oxygen gas will be consumed when 950g of glucose is oxidized completely.