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givi [52]
3 years ago
9

2. Hydrocarbons consist of carbon and hydrogen. 5.7 g of a hydrocarbon contains 4.8 g of carbon.

Chemistry
1 answer:
zaharov [31]3 years ago
7 0

Answer:

The molecular formula of the hydrocarbon compound is C₈H₁₈

Explanation:

The type of chemical substance in the question = Hydrocarbon

The mass of the hydrocarbon = 5.7 g

The mass of carbon in the given hydrocarbon (sample) = 4.8 g

The relative molecular mas of the hydrocarbon = 114 g

Therefore, the mass of hydrogen in the sample, H = 5.7 g - 4.8 g = 0.9 g

The ratio of the mass of carbon in the sample = 4.8/5.7

The ratio of the mass of hydrogen in the sample = 0.9/5.7

By the law of constant composition, we have;

The mass of carbon in a mole of the hydrocarbon = (4.8/5.7) × 114 g = 96 g

The molar mass of carbon, C = 12 g/mol

The number of moles of carbon in a mole of the hydrocarbon, 'n₁', is given as follows;

n₁ = 96 g/12 g = 8 moles

The mass of hydrogen in a mole of the hydrocarbon = (0.9/5.7) × 114 g = 18 g

The molar mass of hydrogen, H ≈ 1 g/mol

The number of moles of hydrogen in a mole of the hydrocarbon, 'n₂', is given as follows;

n₂ = 48 g/1 g = 18 moles

Therefore, in each each molecule of the compound, we have have 8 atoms of carbon, 'C', and 18 atoms of hydrogen, 'H'

Therefore, we get;

The molecular formula of the hydrocarbon compound is C₈H₁₈ which is the chemical formula for the hydrocarbon also known as octane

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Answer:

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3 0
3 years ago
What is the reaction for CL^2 + 2 KBr —> 2 KCL+Br^2 of 11 grams of potassium bromide?
mylen [45]

Answer:

All the amounts of reactants and products are:

KBr

          11.0g (given)

          0.0924 mol

Cl₂

           0.0462 mol

           3.28g

KCl

           0.0924 mol

           6.89 g

Br₂

         0.0462 mol

         7.39 g

         

Explanation:

<u>1. Balanced chemical equation (given)</u>

   Cl_2+2KBr\rightarrow 2KCl+Br_2

<u>2. Mole ratios</u>

     \dfrac{1molCl_2}{2molKBr}

      \dfrac{2molKCl}{2molKBr}

     \dfrac{1molBr_2}{2molKBr}

<u />

<u>3. Molar masses</u>

  • Molar mass Cl₂: 70.906g/mol
  • Molar mass KBr: 119.002 g/mol
  • Molar mass KCl: 74.5513 g/mol
  • Molar mass KBr: 159.808 g/mol

<u>4. Convert 11 grams of potassium bromide to moles:</u>

  • #moles = mass in grams / molar mass
  • #mol KBr = 11g / 119.002g/mol = 0.092435mol KBr

<u>5. Use the mole ratios to find the amounts of Cl₂, KCl, and Br₂</u>

a) Cl₂

       \dfrac{1molCl_2}{2molKBr}\times 0.092435molKBr=0.0462molCl_2

        0.0462175molCl_2\times 70.906g/molCl_2=3.28gCl_2

b) KCl

       \dfrac{2molKCl}{2molKBr}\times0.092435molKBr=0.0924molKCl

      0.092435molKCl\times 74.5513g/molKCl=6.89gKCl

c) Br₂

       

         \dfrac{1molBr_2}{2molKBr}\times 0.092435molKBr=0.0462molBr_2

         0.0462175molBr_2\times 159.808g/molBr_2=7.39gBr2

The final calculations are rounded to 3 sginificant figures.

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