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Ammonium carbonate decomposes upon heating according to

BaLLatris [955]

The total volume of the gas produced, given the data from the question is 11.84 L

<h3>How to determine the mole of (NH₄)₂CO₃</h3>

Mass of (NH₄)₂CO₃ = 11.83 g
Molar mass of (NH₄)₂CO₃ = 96 g/mol
Mole of (NH₄)₂CO₃ =?

Mole = mass / molar mass

Mole of (NH₄)₂CO₃ = 11.83 / 96

Mole of (NH₄)₂CO₃ = 0.123 mole

<h3>How to determine the volume of NH₃</h3>

Balanced equation

(NH₄)₂CO₃ -> 2NH₃(g) + CO₂(g) + H₂O(g)

From the balanced equation above,

1 mole of (NH₄)₂CO₃ decomposed to produce 2 moles of NH₃

Therefore,

0.123 mole of (NH₄)₂CO₃ will decompose to produce = 0.123 × 2 = 0.246 mole of NH₃

Thus, we can determine the volume of NH₃ produced by using the ideal gas equation as follow:

Temperature (T) = 22 °C = 22 + 273 = 295 K
Pressure (P) = 1.02 bar
Gas constant (R) = 0.08314 bar.L/Kmol
Number of mole (n) = 0.246 moles
Volume of NH₃ (V) =?

PV = nRT

Divide both side by P

V = nRT / P

V = (0.246 × 0.08314 × 295) / 1.02

Volume of NH₃ = 5.92 L

<h3>How to determine the volume of CO₂</h3>

Balanced equation

(NH₄)₂CO₃ -> 2NH₃(g) + CO₂(g) + H₂O(g)

From the balanced equation above,

1 mole of (NH₄)₂CO₃ decomposed to produce 1 moles of CO₂

Therefore,

0.123 mole of (NH₄)₂CO₃ will also decompose to produce = 0.123 mole of CO₂

Thus, we can determine the volume of CO₂ produced by using the ideal gas equation as follow:

Temperature (T) = 22 °C = 22 + 273 = 295 K
Pressure (P) = 1.02 bar
Gas constant (R) = 0.08314 bar.L/Kmol
Number of mole (n) = 0.123 moles
Volume of CO₂ (V) =?

PV = nRT

Divide both side by P

V = nRT / P

V = (0.123 × 0.08314 × 295) / 1.02

Volume of CO₂ = 2.96 L

<h3>How to determine the volume of H₂O</h3>

Balanced equation

(NH₄)₂CO₃ -> 2NH₃(g) + CO₂(g) + H₂O(g)

From the balanced equation above,

1 mole of (NH₄)₂CO₃ decomposed to produce 1 moles of H₂O

Therefore,

0.123 mole of (NH₄)₂CO₃ will also decompose to produce = 0.123 mole of H₂O

Thus, we can determine the volume of H₂O produced by using the ideal gas equation as follow:

Temperature (T) = 22 °C = 22 + 273 = 295 K
Pressure (P) = 1.02 bar
Gas constant (R) = 0.08314 bar.L/Kmol
Number of mole (n) = 0.123 moles
Volume of H₂O (V) =?

PV = nRT

Divide both side by P

V = nRT / P

V = (0.123 × 0.08314 × 295) / 1.02

Volume of H₂O = 2.96 L

<h3>How to determine the total volume of gas produced</h3>

Volume of NH₃ = 5.92 L
Volume of CO₂ = 2.96 L
Volume of H₂O = 2.96 L
Total volume of gas =?

Total volume = 5.92 + 2.96 + 2.96

Total volume of gas = 11.84 L

Learn more about ideal gas equation:

brainly.com/question/4147359

Learn more about stoichiometry:

brainly.com/question/14735801

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7 0

2 years ago

1A: Consider these compounds:

Setler [38]

Solution :

Compound Ksp

$PbF_2$ $3.3 \times 10^{-8}$

$Ni(CN)_2$ $3 \times 10^{-23}$

FeS $8 \times 10^{-19}$

$CaSO_4$ $4.93 \times 10^{-5}$

$Mg(OH)_2$ $5.61 \times 10^{-12}$

Ksp of $$Ni(CN)_2$ and both compounds dissociate the same way. Hence $Mg(OH)_2$ is more soluble than $(B). \ Ni(CN)_2$

$Mg(OH)_2$ is less soluble than $(A). \ \ PbF_2 \ ()Ksp \ PbF_2 > Ksp \ \text{ of } \ Mg(OH)_2$

It is not possible to determine CD - $FeS \text{ or} \ CaSO_4$ is more or less soluble than $Mg(OH)_2$ as though they have a different Ksp values their molecular dissociation is also different and they may have a close solubility values.

$Ni(OH)_2$ can be directly compared with PbS, $AlPO_4, MnS$

$\text{For } \ Ni(OH)_2$

$AB_2(s) \rightarrow A^{2+} + 2B^{-}$

$Ni(OH)_2(s) \rightarrow Ni^{2+} + 2OH^-$

100

1-s s 2s

Ksp = $[A2+][B-]^2 = s \times (2s)^2 = 4s^3$

Hence they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.

For Silver Chloride

$AB(s) \rightarrow A^{x+}+B^{x-}$

$AgCl(s) \rightarrow Ag^+ + Cl^-$

1 0 0

1 - s s s

Ksp $=[A^{x+}][B^{x-}]=s \times s = s^2$

Hence, they can be directly compared by Ksp values, smaller the Ksp, smaller the solubility.

4 0

3 years ago

Deffinition of density

Bas_tet [7]

Solidity, solidness, denseness, thickness, substance, mass......

4 0

3 years ago

Read 2 more answers

Soil texture is based on?

blsea [12.9K]

It is based on the water underground

3 0

3 years ago

A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the

tino4ka555 [31]

Answer:

2.1 × 10⁻¹ M

2.0 × 10⁻¹ m

Explanation:

Molarity

The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:

3.9 g × (1 mol/93.13 g) = 0.042 mol

The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:

M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M

Molality

The moles of solute are 0.042 mol.

The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:

200 mL × 1.05 g/mL = 210 g = 0.210 kg

The molality of aniline is:

m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m

5 0

3 years ago