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katen-ka-za [31]
3 years ago
11

How many moles of glucose, C H O , can be "burned" biologically when 18.2 mol of oxygen is available? C H O (s) + 6O (g) --->

6CO (g) + 6H O(l) please show all the work
Chemistry
2 answers:
eimsori [14]3 years ago
8 0
Answer: 18.2


Explanation:

1) Chemical equation: <span>C₆ H₁₂ O₆ (s) + 6O₂ (g) ---> 6CO₂ (g) + 6H₂O(l)</span>

2) Mole ratio: 6 moles <span>C₆ H₁₂ O₆ : 6 mol O₂</span>

6:6 is the same that 1:1. This is, each molecule of <span>C₆ H₁₂ O₆ is burned with one mole of O₂.</span>

3) Conclusion:

Therefore, 18.2 moles of <span>C₆ H₁₂ O₆ can be burned with 18.2 moles of O₂</span>


Feliz [49]3 years ago
5 0
Answer:
              547.7 g of C₆H₁₂O₆

Solution:
               The balance chemical equation is as follow,

                           C₆H₁₂O₆  +  6 O₂    →    6 CO₂  +  6 H₂O

According to equation,

                         6 moles of O₂ burns  =  180.56 g of C₆H₁₂O₆
So,
                18.2 moles of O₂ will burn  =  X g of C₆H₁₂O₆

Solving for X,
                      X  =  (18.2 mol × 180.56 g) ÷ 6 mol

                      X  =  547.7 g of C₆H₁₂O₆
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The rate constant for the second-order reaction 2NOBr(g) ¡ 2NO(g) 1 Br2(g) is 0.80/M ? s at 108C. (a) Starting with a concentrat
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Explanation:

Hello,

(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:

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