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otez555 [7]
3 years ago
8

Sulfuric acid is essential to dozens of important industries from steelmaking to plastics and pharmaceuticals. More sulfuric aci

d is made than any other industrial chemical, and world production exceeds 2.0 x 10 kg per year. The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces 4.4 kg of solid sulfur and 6.90 atm of oxygen gas at 950. °C into an evacuated 50.0 L tank. The engineer believes K-0.71 for the reaction at this temperature. Calculate the mass of solid sulfur he expects to be consumed when the reaction reaches equilibrium. Round your answer to 2 significant digits. Note for advanced students: the engineer may be mistaken in his belief about the value of Kp, and the consumption of sulfur you calculate may not be what he actually observes.
Chemistry
1 answer:
Pavlova-9 [17]3 years ago
6 0

Given:

K = 0.71 = Kp

The reaction of sulphur with oxygen is

                            S(s)   + O2(g)  ---> SO2(g)

initial Pressure                   6.90         0

Change                                -x            +x

Equilibrium                     6.90-x          x

Kp = pSO2 / pO2 = 0.71 = x / (6.90-x)

4.899 - 0.71x  = x

4.899 = 1.71x

x = 2.86 atm = pressure of SO2 formed

temperature = 950 C = 950 + 273.15 K = 1223.15 K

Volume = 50 L

Let us calculate moles of SO2 formed using ideal gas equation as

PV = nRT

R = gas constant = 0.0821 L atm / mol K

putting other values

n = PV / RT = 2.86 X 50 / 1223.15 X 0.0821 = 1.42 moles

Moles of Sulphur required = 1.42 moles

Mass of sulphur required or consumed = moles X atomic mass of sulphur

mass of S = 1.42 X 32 = 45.57 grams or 0.04557 Kg  of sulphur



 


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The formal charges of all nonhydrogen atoms are -1.

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<u>O 7-4 = 3 O Double bond on one H 5-4 = 1</u>

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It will percentage its last valence electron thru a single bond to the terminal oxygen atom. This is in agreement with carbon and hydrogen atoms that each need to form 4 and 1 covalent bonds respectively. because the terminal oxygen atom best has a single covalent bond, it'll have a proper rate of -1.

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Consider the reaction: CH4 + 2O2 = CO2 + 2H2O
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Answer:

The answer to your question is:

a)  80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction                             CH4   +   2O2   ⇒   CO2   +   2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

                               16 g of CH4 ----------------  2(32) g of O2

                               20 g              --------------    x

                               x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .  

                                CH4   +   2O2   ⇒   CO2   +   2H2O

                                15 g         22 g

                                16 g of CH4 ----------------  64 g of O2

                                15 g of CH4  ---------------   x

                               x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

                                 CH4   +   2O2   ⇒   CO2   +   2H2O

                        64 g of O2 ------------------  44 g of CO2

                        22 g of O2 ------------------   x

                        x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________  

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We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

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2 years ago
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