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3 years ago

Please help me to solve this question, thank you so much!!!!

Chemistry

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V125BC [204]3 years ago

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Answer:

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What is the ph of a solution of 0.700 m kh2po4, potassium dihydrogen phosphate?

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Molal or molar? There is a difference

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3 years ago

What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2

atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

$Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}$

$Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}$

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

$Mole of AlCl_3 = Mole of Al * Mole ratio$

$Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}$

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = $133.34 \frac{g}{mol}$

Mass of AlCl3 can be calculated using mole formula as:

$1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}$

Multiplying both side by $133.34 \frac{g}{mol}$

$1.32 mole * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34 \frac{g}{mol}$

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

$Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}$

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

$Mole of AlCl_3 = mole of Cl_2 * mole ratio$

$Mole of AlCl_3 = 0.55 mole * \frac{2}{3}$

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

$0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}$

Multiplying both side by

$133.34 \frac{g}{mol}$

$0.367 mol of AlCl_3 * 133.34 \frac{g}{mol} = \frac{mass(g)}{133.34\frac{g}{mol}} * 133.34 \frac{g}{mol}$

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

7 0

3 years ago

Read 2 more answers

Name three parts of a plant cell that are not found in an animal cell

lions [1.4K]

The three parts are vacuole , cell wall , chloroplast

4 0

3 years ago

Atom X has 50 nucleons and a binding energy of 4.2 10^11 J. Atom Z has 80 nucleons and a binding energy of 8.4 10^11 J. Which at

77julia77 [94]

Answer:

E(Z) > E(X)

Explanation:

X => 4.2 x 10¹¹J/50 Nucleons = 8.4 x 10⁹ J/Nu

Z => 8.4 x 10¹¹J/80 Nucleons = 1.1 x 10¹⁰ J/Nu

E(Z)1.1 x 10¹¹J/Nu > E(X)8.4 x 10⁹J/Nu

3 0

4 years ago

0.60 atm of SO3 and 0.30 atm of SO2are placed in a container and the system is allowed to reach equilibrium. Calculate the press

frozen [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The pressure is $[O_2] = 4.8 *10^{-5} \ atm$

Explanation:

From the question we are told that

The pressure of $SO_3$ is $[SO_3 ] = 0.63 \ atm$

The pressure of $SO_2$ is $[SO_ 2] = 0.30 \ atm$

The equilibrium constant is $K_p = 1.2 *10^{-5}$

The reaction is

$2SO_3 _{(g)}$ ⇔ $2SO_2_{(g)} + O_2 _{(g)}$

Generally the equilibrium constant is mathematically represented as

$K_p = \frac{(SO_2)^2 * (O_2)}{(SO_3)^2}$

=> $[O_2] = \frac{k_p * [SO_3] ^2 }{[SO_2]^2}$

substituting values

$[O_2] = \frac{1.2 *10^{-5} * 0.60 ^2 }{0.30^2}$

$[O_2] = 4.8 *10^{-5} \ atm$

8 0

3 years ago

Please help me to solve this question, thank you so much!!!!​

Please help me to solve this question, thank you so much!!!!