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Fittoniya [83]
3 years ago
11

How many liters of oxygen gas can be produced if 28.7 grams of water decomposes at 294 Kelvin and 0.986 atmospheres? Show all of

the work used to solve this problem. . . 2 H2O (l) --->2 H2 (g) + O2 (g)
Chemistry
1 answer:
Shtirlitz [24]3 years ago
5 0
2 H₂O (l) → 2 H₂ (g) + O₂ (g) 

Molar mass water = 1.01 x 2 + 16.00 = 18.02 g/mol 

Number of mol water decomposed = 28.7 g H₂O x [1 mol / 18.02g] = 1.59 mol H₂O 

<span>From the balanced equation 2 mol H₂O decomposes to 2 mol H₂ and 1 mol O₂ </span>

so the mole ratio water : oxygen = 2 : 1 

and number of mol O₂ produced = ½ x 1.59 = 0.796 mol O₂ 


The ideal gas law is PV = nRT 

so V = nRT/P 

P = 0.986 atm 

V = ? 

n = 0.796 

R = 0.0821 L atm K⁻¹ mol⁻¹ 

T = 294K 

V = 0.796 x 0.0821 x 294 / 0.986 

V = 19.5 L 

<span>So 19.5 L O₂ gas are produced

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
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In one step in the synthesis of the insecticide Sevin, naphthol reacts with phosgene as shown.
Serhud [2]

Answer:

Explanation:

Chemical equation:

C₁₀H₈O + COCl₂  → C₁₁H₇O₂Cl + HCl

A. How many kilograms of C₁₁H₇O₂Cl form from 2.5×10*2 kg of naphthol?

Given data:

Mass of naphthol = 2.5 ×10² kg ( 250×1000 = 250000 g)

Mass of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 250000 g/ 144.17 g/mol

Number of moles of naphthol = 1734.1 mol

Now we will compare the moles of naphthol with C₁₁H₇O₂Cl.

                     C₁₀H₈O       :         C₁₁H₇O₂Cl

                        1               :                1

                       1734.1         :             1734.1

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 1734.1 mol × 206.5 g/mol

Mass = 358091.65 g

Gram to kilogram:

1 kg×358091.65 g/ 1000 g  = 358.1 kg

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

Given data:

Mass of naphthol = 100 g

Mass of COCl₂ = 100 g

Theoretical yield of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 100 g/ 144.17 g/mol

Number of moles of naphthol = 0.694 mol

Number of moles of phosgene:

Number of moles  = mass/ molar mass

Number of moles =  100 g/ 99 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of naphthol and phosgene with C₁₁H₇O₂Cl.

                     C₁₀H₈O        :         C₁₁H₇O₂Cl

                        1                :                1

                       0.694        :              0.694

                    COCl₂          :             C₁₁H₇O₂Cl

                        1                :                1

                       1.0              :              1.0

The number of moles of C₁₁H₇O₂Cl produced by C₁₀H₈O are less so it will limiting reactant and limit the yield of  C₁₁H₇O₂Cl.

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass =  0.694 mol × 206.5 g/mol

Mass = 143.3 g

Theoretical yield  =  143.3 g

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Given data:

Actual yield of C₁₁H₇O₂Cl = 118 g

Theoretical yield = 143.3 g

Percent yield = ?

Solution:

Formula :

Percent yield = actual yield / theoretical yield × 100

Now we will put the values in formula.

Percent yield = 118 g/ 143.3 g × 100

Percent yield = 0.82 × 100

Percent yield = 82%

5 0
3 years ago
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