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charle [14.2K]
2 years ago
13

Balance Reaction Al(s) + HCl(aq) —> AlCl3 + H2

Chemistry
2 answers:
musickatia [10]2 years ago
3 0

Answer:

2 Al + 6 HCl ⇒ 2 AlCl₃ + 3 H₂

storchak [24]2 years ago
3 0
Yeaaa it’s right the top of me
You might be interested in
For the reaction 2 s (s) + 3 o2 (g) → 2 so3 (g), how much so3 can be produced from 4 g o2 and excess s? 1. 0.25 mol so3 2. 0.13
igor_vitrenko [27]
The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
8 0
3 years ago
Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

5 0
3 years ago
). The molar mass of an organic acid, a compound composed of carbon, hydrogen, and oxygen, is 194.14 g/mol. Combustion of a 1.50
Nikitich [7]

Answer:

The empirical formula is C₆H₁₀O₇.

Step-by-step explanation:

1. Calculate the masses of C, H, and O from the masses given.

Mass of C =  2.0402 g CO₂ × (12.01 g C/44.01 g CO₂) = 0.5568  g C

Mass of H = 0.6955 g H₂O  × (2.016 H/18.02 g H₂O)  = 0.077 81 g H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.500 – 0.5568 – 0.077 81) g = 0.8654 g O

=====

2. Convert these masses to moles.

Moles  C = 0.5568  × 1/12.01  = 0.046 36

Moles H = 0.077 81 × 1/1.008 = 0.077 19

Moles O = 0.8654   × 1/16.00 = 0.054 09

=====

3. Find the molar ratios.

Moles  C = 0.046 36/0.046 36 = 1

Moles H = 0.077 19/0.046 36   = 1.665

Moles O = 0.054 09/0.046 36 = 1.167

======

4. Multiply the ratios by a number to make them close to integers

C  = 1        × 6 = 6

H = 1.665 × 6 = 9.991

O = 1.167 × 6  = 7.001

=====

5. Round the ratios to integers

C:H:O =6:10:7

=====

6. Write the empirical formula

The empirical formula is C₆H₁₀O₇.

=======

7. Calculate the empirical formula mass

C₆H₁₀O₇ = 6×12.01 + 10×1.008 + 7×16.00

C₆H₁₀O₇ = 72.01 + 10.08+ 112.0

C₆H₁₀O₇ = 194.09

=====

8. Divide the molecular mass by the empirical formula mass.  

MM/EFM = 194.14/194.09 = 1.000 ≈ 1

=====

9. Determine the molecular formula

MF = (EF)ₙ = (C₆H₁₀O₇)₁ = C₆H₁₀O₇

7 0
3 years ago
Some common substances used in the laboratory are listed in the table. the chemical formulas of the substances are also listed b
Helga [31]
Its 5 i think

but dont take my word on it

7 0
3 years ago
The normal freezing point of a certain liquid Xis-7.30°C but when l02. g of iron(III) chloride (FeCl3) are dissolved in 650. g o
IRISSAK [1]

Answer:

2.7 °C.kg/mol

Explanation:

Step 1: Calculate the freezing point depression (ΔT)

The normal freezing point of a certain liquid X is-7.30°C and the solution freezes at -9.9°C instead. The freezing point depression is:

ΔT = -7.30 °C - (-9.9 °C) = 2.6 °C

Step 2: Calculate the molality of the solution (b)

We will use the following expression.

b = mass of solute / molar mass of solute × kilograms of solvent

b = 102. g / (162.2 g/mol) × 0.650 kg = 0.967 mol/kg

Step 3: Calculate the molal freezing point depression constant Kf of X

Freezing point depression is a colligative property. It can be calculated using the following expression.

ΔT = Kf × b

Kf = ΔT / b

Kf = 2.6 °C / (0.967 mol/kg) = 2.7 °C.kg/mol

7 0
3 years ago
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