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leonid [27]
3 years ago
11

If 1.0 mole of CH4 and 2.0 moles of Cl2 are used in the reaction CH4 + 4Cl2 => CCl4 + 4HCl then which of these statements is

TRUE?
A. All of the CH4 will be used up.
B. The theoretical yield of CCl4 is 1.0 mole.
C. Some Cl2 will be left over.
D. More CH4 should be added to ensure all the Cl2 will be used.
E. The theoretical yield of HCl is 2.0 moles.
Chemistry
1 answer:
nikklg [1K]3 years ago
4 0

Answer:

B,C,D

Explanation:

The yield of CCl4 depends on the amount of CH4 in a 1:1 ratio. The amount of Cl2 is twice that of CH4 hence some must be left over. To ensure that all the Cl2 is used up, more CH4 must added to the system.

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Explanation:

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8 0
3 years ago
Menthol, the substance we can smell in mentholated cough drops, is composed of c, h, and o. a 9.045×10−2 −mg sample of menthol i
Ket [755]

Answer:

            Empirical Formula  =  C₁₀H₂₀O

Solution:

Data Given:

                      Mass of Menthol  =  9.045 × 10⁻² mg  =  9.045 × 10⁻⁵ g

                      Mass of CO₂  =  0.2546 mg  =  0.0002546 g

                      Mass of H₂O  =  0.1043 mg  =  0.0001043 g

Step 1: Calculate %age of Elements as;

                      %C  =  (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

                      %C  =  (0.0002546 ÷ 9.045 × 10⁻⁵) × (12 ÷ 44) × 100

                      %C  =  (2.814) × (12 ÷ 44) × 100

                      %C  =  2.814 × 0.2727 × 100

                      %C  =  76.73 %


                      %H  =  (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.0001043 ÷ 9.045 × 10⁻⁵) × (2.02 ÷ 18.02) × 100

                      %H  =  (1.153) × (2.02 ÷ 18.02) × 100

                      %H  =  1.153 × 0.1120 × 100

                     %H  =  12.91 %


                      %O  =  100% - (%C + %H)

                      %O  =  100% - (76.73% + 12.91%)

                      %O  =  100% - 89.64%

                     %O  =  10.36 %

Step 2: Calculate Moles of each Element;

                      Moles of C  =  %C ÷ At.Mass of C

                      Moles of C  = 76.73 ÷ 12.01

                     Moles of C  =  6.3888 mol


                      Moles of H  =  %H ÷ At.Mass of H

                      Moles of H  = 12.91 ÷ 1.01

                      Moles of H  =  12.7821 mol


                      Moles of O  =  %O ÷ At.Mass of O

                      Moles of O  = 10.36 ÷ 16.0

                      Moles of O  =  0.6475 mol

Step 3: Find out mole ratio and simplify it;

                C                                        H                                     O

            6.3888                              12.7821                            0.6475

     6.3888/0.6475                  12.7821/0.6475                 0.6475/0.6475

               9.86                                   19.74                                   1

             ≈ 10                                      ≈ 20                                     1

Result:

         Empirical Formula  =  C₁₀H₂₀O₁

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If someone trys to lur u into a car with candy...say NO. and run away
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