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4 years ago

FeSO4 • 7H2O

Chemistry

2 answers:

ziro4ka [17]4 years ago

8 0

Answer:

The percent by mass of water in the hydrate is 45.4%

The percent by mass of the anhydrous salt in the hydrate is 54.6%

Explanation:

Mrac [35]4 years ago

4 0

Answer:

1. Percent by mass of H₂O = 45.3%

2. Percent by mass of anhydrous Salt (FeSO₄) = 54.7 %

Explanation:

Data Given

Formula of the Molecule = FeSO₄ • 7H₂O

% by mass water (H₂O) = ?

% by mass FeSO₄ = ?

________________________________________

> First of all find the atomic masses of each component in a molecule

For H₂O atomic masses are given below

H = 1 g/mol

O = 16 g/mol

> Then find the total mass of H₂O in haydrated salt

7H₂O = 7 (2x1 +1x16) g/mol

7H₂O = 7 (2+16) g/mol

7H₂O = 7 (18) g/mol

7H₂O = 126 g/mol

> find total Molar Mass of Molecule:

Molar Mass of FeSO₄ • 7H₂O = [56+ 32 + 16x4] + 7(2+16)

Molar Mass of FeSO₄ • 7H₂O = [56+ 32 + 64] + 126

Molar Mass of FeSO₄ • 7H₂O = 278

Now to find the mass by percent of H₂O

Formula used to find the mass by percent of a component

Percent composition of H₂O = mass of H₂O in Molecule / molar mass of FeSO₄ • 7H₂O x 100%

Put the values

Percent by mass of H₂O = 126 (g/mol) / 278 (g/mol) x 100%

Percent by mass of H₂O = 0.4532 x 100%

Percent by mass of H₂O = 45.3%

_______________________________________________

> First of all find the atomic masses of each component in a molecule

For anhydrus salt (FeSO₄) atomic masses are given below

Fe = 56 g/mol

S= 32 g/mol

O = 16 g/mol

> Then find the total mass of FeSO₄ in haydrated salt

FeSO₄ = (56x1 + 32 + 4x16) g/mol

FeSO₄ = ( 56 + 32 + 64) g/mol

FeSO₄ = 152 g/mol

> find total Molar Mass of Molecule:

Molar Mass of FeSO₄ • 7H₂O = [56+ 32 + 16x4] + 7(2+16)

Molar Mass of FeSO₄ • 7H₂O = [56+ 32 + 64] + 126

Molar Mass of FeSO₄ • 7H₂O = 278

Now to find the mass by percent of FeSO₄

Formula used to find the mass by percent of a component

Percent composition of FeSO₄ = mass of FeSO₄ in Molecule / molar mass of FeSO₄ • 7H₂O x 100%

Put the values

Percent by mass of FeSO₄ = 152 (g/mol) / 278 (g/mol) x 100%

Percent by mass of FeSO₄ = 0.547 x 100%

Percent by mass of FeSO₄ = 54.7 %

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How many liters of CO2 gas can be produced at 30.0 °C and 1.50 atm from the reaction of 5.00 mol of C3H8 and an excess of O2 acc

lbvjy [14]

Answer:

249 L

Explanation:

Step 1: Write the balanced equation

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

Step 2: Calculate the moles of CO₂ produced from 5.00 moles of C₃H₈

The molar ratio of C₃H₈ to CO₂ is 1:3. The moles of CO₂ produced are 3/1 × 5.00 mol = 15.0 mol

Step 3: Convert "30.0°C" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 30.0°C + 273.15 = 303.2 K

Step 4: Calculate the volume of carbon dioxide

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 15.0 mol × 0.0821 atm.L/mol.K × 303.2 K/1.50 atm

V = 249 L

5 0

3 years ago

The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p

Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

$\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}$

(ii) Solve the relationship

Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

56x + 18(7.5 - x) = 267.3

56x + 135 - 18x = 267.3

38x = 132.3

x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

4 0

4 years ago

Type of reaction in NaOH + KNO3 → NaNO3 + KOH

Lesechka [4]

Answer:

It is a combustion reaction

Explanation:

6 0

3 years ago

Help me with this work plz plz

storchak [24]

Answer:

a -4

b - 8

c - 5

d 2

Explanation:

Significant Figures: The number of digits used to express a measured or calculated quantity. By using significant figures, we can show how precise a number is. ... Accuracy: Refers to how closely individual measurements agree with the correct or true value.

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2 years ago

The compound HNO3 can be described as an

xz_007 [3.2K]

The answer is (1) Arrhenius acid and an electrolyte. The HNO3 is a H+ ion donor and OH- receptor. Electrolyte means the compound will release ion when dissolving in suitable ionized solution.

6 0

3 years ago