Answer :

you need to find the mole of glucose by using the formula n = m/Mr. Once you have found the mole you need to follow the stoichiometric process by unknown/known using the co-efficients. Then you need to multiple your answer by the known mole of glucose whoch you have previously calculated. After that you should get the mole of oxygen then you just need to transpose the equation to get m = n×Mr, substitute ypur answers in, remebering you can find your Mr via the period table, you have your answer
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Answer:

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

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?

[tex]grams of oxygen=\frac{950 grams of glucose*192 grams of oxygen}{180 grams of glucose}[/tex]

grams of oxygen=1,013

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

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