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2 years ago

AlCl3 and NaOH Balanced Equation Total Ionic Equation Net Ionic Equation

Chemistry

1 answer:

Sindrei [870]2 years ago

7 0

Answer:

Here's what I get

Explanation:

1. In dilute NaOH

(a) Molecular equation

AlCl₃(aq) + 3NaOH(aq) ⟶ Al(OH)₃(s) + 3NaCl(aq)

(b) Ionic equation

You write the molecular formulas for solids, and you write the soluble ionic substances as ions.

Al³⁺(aq) + 3Cl⁻(aq) + 3Na⁺(aq) + 3OH⁻(aq) ⟶ Al(OH)₃(s) + 3Na⁺(aq) + 3Cl⁻(aq)

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

Al³⁺(aq) + 3Cl⁻(aq) + 3Na⁺(aq) + 3OH⁻(aq) ⟶ Al(OH)₃(s) + 3Na⁺(aq) + 3Cl⁻(aq)

The net ionic equation is

Al³⁺(aq) + 3OH⁻(aq) ⟶ Al(OH)₃(s)

2. In excess NaOH

The aluminium hydroxide reacts with excess hydroxide to form sodium tetrahydroxoaluminate(III).

(a) Molecular equation

AlCl₃(aq) + 4NaOH(aq) ⟶ NaAl(OH)₄(aq) + 3NaCl(aq)

(b) Ionic equation

Al³⁺(aq) + 3Cl⁻(aq) + 4Na⁺(aq) + 4OH⁻(aq) ⟶ Na⁺Al(OH)₄⁻(aq) + 3Na⁺(aq) + 3Cl⁻(aq)

(c) Net ionic equation

Al³⁺(aq) + 4OH⁻(aq) ⟶ Al(OH)₄⁻(aq)

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How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group

OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

P= 1.00 atm
V= 150 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 $\frac{atm*L}{mol*K}$ *273 K

Solving:

$n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}$

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

$mass=\frac{6.69 moles*70.9 grams}{1 mole}$

mass= 474.321 grams ≅ 474 grams

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

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