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Fittoniya [83]
3 years ago
6

You have arrived on a planet where S2, not O2, is the gas used as fuel in combustion reactions. When gaseous propane (C3H8) is b

urned in S2, H2S and carbon disulfide gases are produced. If 200.0 grams of propane and 75.0 grams of S2 were reacted, what is the theoretical yield of carbon disulfide. Include balance equations
Chemistry
1 answer:
Debora [2.8K]3 years ago
8 0

Answer:

Mass = 53.45 g

Explanation:

Given data:

Mass of propane = 200 g

Mass of S₂ = 75 g

Theoretical yield /Mass of CS₂ = ?

Solution:

Balanced Chemical equation:

C₃H₈ + 5S₂      →    4H₂S + 3CS₂

Number of moles of propane:

Number of moles = mass / molar mass

Number of moles = 200 g/  44.1 g/mol

Number of moles = 4.54 mol

Number of moles of S₂:

Number of moles = mass / molar mass

Number of moles = 75 g/  64.14 g/mol

Number of moles = 1.17 mol

Now we will compare the moles of carbon disulfide with both reactant.

                        S₂             :           CS₂

                         5              :            3

                        1.17           :          3/5×1.17 = 0.702

                       C₃H₈          :            CS₂

                           1             :               3

                        4.54          :              3×4.54 = 13.62 mol

Number of moles of CS₂ produced by S₂ are less so it will limiting reactant and limit the yield of carbon disulfide.

Theoretical yield of carbon disulfide.

Mass = number of moles ×molar mass

Mass = 0.702 mol × 76.14 g/mol

Mass = 53.45 g

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Data

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Process

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                      3.268 g of CO₂  --------    x

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                      1.672 g of H₂O ------------  x

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Moles of Oxygen

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                             0.299 g of O    -----------------  x

                             x = (0.299 x 1) / 16

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Hydrogen     0.186/ 0.019 = 9.7 = 10

Oxygen         0.019/ 0.019 = 1

5.- Write the empirical formula

                              C₄H₁₀O                  

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