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Lelechka [254]
3 years ago
11

How many grams of Bromine are required to react completely with 11.95 grams aluminum chloride? AlCl3+Br2-->AlBr3+Cl2

Chemistry
2 answers:
liberstina [14]3 years ago
6 0
First have to balance the equation of  AlCl3 + Br2 ---> Br3 + Cl2:

                                  2AlCl3 + 3Br3----> 2AlBr3 + 2Cl2

                           1 mol AlCl3           3mol Br2         159.8 g Br2                         11.95g AlCl3   x  ---------------      x    -------------     x  --------------- =  21.48g Br2
                          133.33 g AlCl3      2 mol AlCl3       1 mol Br2


(133.33 g = molar mass of AlCl3       3:2 ratio   
159.8 g = molar mass of Br2)

Answer: 21.48 g Br2 required 
zmey [24]3 years ago
5 0
Molar mass:
 
( Br₂ ) = 159.80 g/mol

AlCl₃ = 133.34 g/mol

<span>2 AlCl</span>₃<span> + 3 Br</span>₂<span> = 2 AlBr</span>₃<span> + 3 Cl</span>₂

2 x ( 133.34) g ------------ 3 x ( 159.80) g
11.95 ------------------------ ( mass Bromine )

mass Bromine = 11.95 x 3 x 159.80 / 2 x 133.34

mass Bromine = 5728.83 / 266.68

mass Bromine = 21.48 g

hope this helps!


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vfiekz [6]

Answer:

P1 =4 atm

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P2=1 atm

T2=?

According to Gay-Lussac's Law or Third Gas Law,

P1T2=P2T1

4×T2=1×20

T2= 20/4

T2= 5°C

Answer At 5°C temperature does a gas at 1.00 atm !

7 0
3 years ago
1. A student needs to make 80mL of a 4.5 x 10^-3 M solution of H_3PO_4 (mm = 98) from solid. Describe how the student would do t
slamgirl [31]

Answer:

1. 35 mg of H₃PO₄

2. 27 mol AlF₃; 82 mol F⁻

3. 300 mL of stock solution.

Explanation:

1. Preparing a solution of known molar concentration

Data:

V = 80 mL

c = 4.5 × 10⁻³ mol·L⁻¹

Calculations:

(a) Moles of H₃PO₄

Molar concentration = moles of solute/litres of solution

c = n/V

n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol

(b) Mass of H₃PO₄  

moles = mass/molar mass

n = m/MM

m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg

(c) Procedure

Dissolve 35 mg of solid H₃PO₄  in enough water to make 80 mL of solution,

2. Moles of solute.

Data:

V = 4900 mL

c = 5.6 mol·L⁻¹

Calculations:

Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃

Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.

3. Dilution calculation

Data:

V₁= 750 mL; c₁ = 0.80 mol·L⁻¹

V₂ = ?            ; c₂ = 2.0   mol·L⁻¹

Calculation:

V₁c₁ = V₂c₂

V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL

Procedure:

Measure out 300 mL of stock solution. Then add 500  mL of water.

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