What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?

Answers

Answer 1

Answer:  " 13.5 g Al " ;

                    →  that is:  "13.5 grams of aluminum."

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Explanation:

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Note: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

 The reactants:  "aluminum (Al) " ;  and "chlorine (Cl) " ;  and:

 The product:    "aluminum choloride (AlCl₃) " .

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The "balanced chemical equation" is:

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        2 Al   +   3 Cl₂   →   2 AlCl₃   ;

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Note: The molecular weight of "aluminum (Al)" is:   " 26.98 g /mol " .

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So:  We call solve using a technique known as:  "dimensional analysis" :

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  0.500 mol AlCl₃ * [tex](\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?[/tex]

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Note:  The units of "mol AlCl₃" cancel out to "1' ; and:

          The  units of "mol Al" cancel out to "1" ; and we are left with:

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 " [tex]\frac{(0.500 * 2 * 26.98)}{2}[/tex]   g Al ["grams of aluminum"] ;

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Note: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

 →  (0.500 * 26.98) g Al ;

    = 13.49 g Al ;

         →  Round to 3 (Three) significant figures;

         →  Since:  "0.500" has 3 (Three) significant figures:

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   =  13.5 g Al ; that is:  "13.5 grams of aluminum."

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 Hope this is helpful!  

      Best wishes to you in your academic pursuits—and within the "Brainly" community!

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a 10 ml sample of borax- borate equilibrium solution reacts completely with 23.33 ml of a 0.1006 M HCl solution. calculate the concentration of borate and then the Ksp for borax​

Answers

Answer:

See answer below

Explanation:

To do this, you need to write down the chemical reaction giving place here.

As this is to explain the thermodynamics of the borax-borate reaction, the general reaction is as follow:

Na₂B₄O₅(OH)₄.8H₂O(s) <---------> 2Na⁺(aq) + B₄O₅(OH)₄²⁻(aq) + 8H₂O(l)

Where:

Na₂B₄O₅(OH)₄.8H₂O(s) = Borax

B₄O₅(OH)₄²⁻(aq) = Borate ion

Now, in order to calculate the concentration of borate, this compound undergoes a titration with HCl. The reaction that takes place here is:

B₄O₅(OH)₄²⁻(aq) + 2HCl + 3H₂O <---------> 4H₃BO₃(aq) + 2Cl⁻(aq)

According to this reaction, it takes 2 moles of HCl to completely react with the borate ion, therefore, the expression to calculate the concentration will be:

2 nHCl = nB₄O₅(OH)₄²⁻

2M₁V₁ = M₂V₂

Replacing the data of concentration and volume of both compounds we have:

2 * 0.1006 * 23.33 = 10M₂

M₂ = 2 * 0.1006 * 23.33 / 10

M₂ = 0.4693 M

This is the concentration of the borate. Now, to calculate the Ksp we need to write the equilibrium reaction taking place with the borax:

Na₂B₄O₅(OH)₄.8H₂O(s) <---------> 2Na⁺(aq) + B₄O₅(OH)₄²⁻(aq) + 8H₂O(l)

Ksp = [B₄O₅(OH)₄²⁻] [Na⁺]²

As we already have the concentration of borax, we can also know the concentration of sodium. If we call concentration as "s" (Stands for solubility), and rewrite the expression of Ksp:

Ksp = s * (2s)²

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Ksp = (0.4693) (2*0.4693)²

Ksp = 0.4693 * 0.881

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Write the balanced equation for the following.

Answers

Answer:

it forms copper hydroxide crystals

Explanation:

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Na+CL2=2NACL is the balanced reaction for the formation of table salt. Given 20 grams of Na and 10 grams of Cl2, which reactant is in excess? How much NaCL should be produced from these amounts of the two reactants?

Answers

Excess reactant : Na

NaCl produced : = 16.497 g

Further explanation

Given

Reaction(balanced)

2Na + Cl₂⇒ 2NaCl

20 g Na

10 g Cl₂

Required

Excess reactant

NaCl produced

Solution

mol Na(Ar = 23 g/mol) :

= 20 : 23 = 0.87

mol Cl₂(MW=71 g/mol):

= 10 : 71 g/mol = 0.141

mol : coefficient :

Na = 0.87 : 2 = 0.435

Cl₂ = 0.141 : 1 = 0.141

Limiting reactant : Cl₂(smaller ratio)

Excess reactant : Na

Mol NaCl based on mol Cl₂, so mol NaCl :

= 2/1 x mol Cl₂

= 2/1 x 0.141

= 0.282

Mass NaCl :

= 0.282 x 58.5 g/mol

= 16.497 g

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