All categories

3 years ago

How many grams of aluminum is produced when 82.4 grams of aluminum chloride

Chemistry

1 answer:

yKpoI14uk [10]3 years ago

6 0

Answer:

16.6 g of Al are produced in the reaction of 82.4 g of AlCl₃

Explanation:

Let's see the decomposition reaction:

2AlCl₃ → 2Al + 3Cl₂

2 moles of aluminum chloride decompose to 2 moles of solid Al and 3 moles of chlorine gas.

We determine the moles of salt:

82.4 g . 1mol/ 133.34g = 0.618 moles

Ratio is 2:2. 2 moles of salt, can produce 2 moles of Al

Then, 0.618 moles of salt must produce 0.618 moles of Al.

Let's convert the moles to mass → 0.618 mol . 26.98g /mol = 16.6 g

You might be interested in

What is electron configuration of oxygen in its excited state

timofeeve [1]

Answer:

$1 {s}^{2} 2 {s}^{2} 2 {p}^{4}$

$2 : 6$

3 0

3 years ago

Read 2 more answers

The equilibrium constant for the formation of ammonia from nitrogen and hydrogen is 1.6 × 102. what is the form of the equilibri

Nimfa-mama [501]

Answer: The expression for equilibrium constant is $\frac{[NH_3]^2}{[H_2]^3[N_2]}$

Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as $k_c$

For a general reaction:

$aA+bB\rightleftharpoons cC+dD$

The equilibrium constant is written as:

$k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

Chemical reaction for the formation of ammonia is:

$N_2+3H_3\rightleftharpoons 2NH_3$

$k_c=1.6\times 10^2$

Expression for $k_c$ is:

$k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}$

$1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}$

8 0

3 years ago

The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t

Andrews [41]

Answer: The experimental van't Hoff factor is 1.21

Explanation:

The expression for the depression in freezing point is given as:

$\Delta T_f=iK_f\times m$

where,

i = van't Hoff factor = ?

$\Delta T_f$ = depression in freezing point = 0.225°C

$K_f$ = Cryoscopic constant = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

$0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21$

Hence, the experimental van't Hoff factor is 1.21

7 0

3 years ago

What is #1 pointing to? a URNA b amino acid chain ribosome С d mRNA

Burka [1]

I farted it tickled my butt cheeks jingled it was hot it was squishy so I did it again

4 0

3 years ago

What is the ph of a solution made by mixing 25.00 ml of 0.100 m hcl with 40.00 ml of 0.100 m koh? assume that the volumes of the

Digiron [165]

Answer: The HCl and KOH will react until one or the other is gone. As you have a larger volume of an equal concentration of HCl, the KOH will go first. moles HCl = 0.04000 L * 0.100 M = 0.00400 moles moles KOH = 0.02500 L * 0.100 M = 0.00250 moles moles HCl left = 0.00400 - 0.00250 = 0.00150 moles Your total volume is now 65.00 mL, so the [HCl] = 0.00150 moles / 0.06500 L = 0.0231 M = [H+] pH = -log [H+] = -log (0.0231) = 1.64

3 0

3 years ago