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mihalych1998 [28]
3 years ago
15

Calculate the energy required to produce 7.00 mol Cl2O7 on the basis of the following balanced equation. 2Cl2(g) + 7O2(g) + 130

kcal --> 2Cl2O7(g) Select one: a. 7.00 kcal b. 65 kcal c. 130 kcal d. 455 kcal
Chemistry
1 answer:
KatRina [158]3 years ago
3 0

Explanation:

As the given chemical reaction equation is as follows.

      2Cl_{2}(g) + 7O_{2}(g) + 130 kcal \rightarrow 2Cl_{2}O_{7}(g)

Also, it is given that for 2 moles the energy required is 130 kcal. This means that energy required for 1 mole is calculated as follows.

                   1 mole = \frac{130 kcal}{2}

                               = 65 kcal

Hence, energy required for 7 moles will be calculated as follows.

              Energy required = 7 \times 65 kcal

                                           = 455 kcal

Thus, we can conclude that energy required to produce 7.00 mol Cl_{2}O_{7} on the basis of given reaction is 455 kcal.

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If the absorbance of a KMnO4 solution of unknown concentration is 0.633, calculate the concentration of KMnO4 in the solution.
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The molar Concentration of KMnO₄ is 0.000219 M

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x is the molar concentration of KMnO_4

y = 4.84E + 03x - 2.26E - 01

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1 year ago
The disturbance of a supersaturated solution will cause
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What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
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Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

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Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

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