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2 years ago

HELP!!!

Chemistry

1 answer:

Evgesh-ka [11]2 years ago

3 0

Answer:

1. Percent yield Trial 1: 93.1 %

2. Leftover reactant for Trial 1: 81.3 g

3. Percent yield Trial 2: 95.1%

4. What ratio of reactants is more efficient: the ratio of the Trial 2: 75g / 50g

Explanation

1) Data (the table is copied for better understanding of the data)

Trial Start. Amount of SiCl₄ Start, Amount of O₂ Actual Yield of SiO₂

1 100 g 100 g 32.96 g

2 75 g 50 g 25.2 g

2) Percent yield for SiO₂ for Trial 1.

a) Balanced chemical equation:

SiCl₄ + O₂ → SiO₂ + 2Cl₂

b) Mole ratio:

1 mol SiCl₄ : 1 mol O₂ : 1 mol SiO₂ : 2Cl₂

c) Limiting reactant:

SiCl₄ (it is given that oxygen is the excess reactant)

d) Convert 100 g of SiCl₄ to moles

moles = mass in grams / molar mass

molar mass of SiCl₄: 169.9 g/mol

moles = 100 g / 169.9 g/mol = 0.589 mol

e) Find moles of SiO₂ using proportions and theoretical mole ratio:

1 mol SiCl₄ / 1 mol SiO₂ = 0.589 mol SiCl₄ / x

⇒ x = 0.589 mol SiO₂

f) Convert 0.589 mol SiO₂ to grams:

mass in grams = number of moles × molar mass

molar mass SiO₂ = 60.08 g/mol

theoretical yield of SiO₂ = 0.589 mol × 60.08 g/mol = 35.4 g

g) Compute percent yield:

percent yield = (actual yield / theoreticl yield) × 100

percent yield = (32.96 g / 35.4 g) × 100 = 93.1%

3) Leftover reactant

a) Convert 100 g of O₂ to moles

moles = mass in grams / molar mass

molar mass of O₂ = 32.0 g/mol

moles of O₂ = 100 g / 32.0 g/mol = 3.13 mol

b) Excess:

3.13 mol - 0.589 mol = 2.54 mol O₂

c) Convert 2.54 mol of O₂ to grams:

mass = number of moles × molar mass = 2.54 mol × 32.0 g/mol = 81.2 g

4) Percent yield for trial 2

a) Convert 75 g of SiCl₄ to moles

moles = mass in grams / molar mass

molar mass of SiCl₄: 169.9 g/mol

moles = 75 g / 169.9 g/mol = 0.441 mol

b) Find moles of SiO₂ using proportions and theoretical mole ratio:

1 mol SiCl₄ / 1 mol SiO₂ = 0.441 mol SiCl₄ / x

⇒ x = 0.441 mol SiO₂

c) Convert 0.441 mol SiO₂ to grams:

mass in grams = number of moles × molar mass

molar mass SiO₂ = 60.08 g/mol

theoretical yield of SiO₂ = 0.441 mol × 60.08 g/mol = 26.5 g

d) Compute the percentage yield:

percentage yield = (actual yield / theoreticl yield) × 100

percentage yield = (25.2 g / 26.5 g) × 100 = 95.1%

e) Conclusion: Since the second trial had a greater percentage yield than the first trial (95.1% vs 93.1 %), the ratio of the Trial 1 is more efficient for the given reaction.

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You have two 500.0 ml aqueous solutions. solution a is a solution of a metal nitrate that is 8.246% nitrogen by mass the ionic c

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1) Answer is: the ionic compound in the solution b is K₂CrO₄ (potassium chromate).

Ionic compound in solution b has two potassiums (oxidation number +1), one chromium (oxidation number +6) and four oxygens. Oxidation number of oxygen is -2 and compound has neutral charge:

2 · (+1) + 6 + x · (-2) = 0.

x = 4; number of oxygen atoms.

2) Answer is: the ionic compound in solution a is AgNO₃ (silver nitrate).

ω(N) = 8.246% ÷ 100%.

ω(N) = 0.08246; mass percentage of nitrogen.

M(MNO₃) = M(N) ÷ ω(N).

M(MNO₃) = 14 g/mol ÷ 0.08246.

M(MNO₃) = 169.8 g/mol; molar mass of metal nitrate.

M(M) = M(MNO₃) - M(N) - 3 · M(O).

M(M) = 169.8 g/mol - 14 g/mol - 3 · 16 g/mol.

M(M) = 107.8 g/mol; atomic mass of metal, this metal is silver (Ag).

3) Balanced chemical reaction:

2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq).

Ionic reaction:

2Ag⁺(aq) + 2NO₃(aq) + 2K⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq).

Net ionic reaction: 2Ag⁺(aq) + CrO₄²⁻(aq) → Ag₂CrO₄(s).

Answer is: the blood-red precipitate is silver chromate (Ag₂CrO₄).

4) m(Ag₂CrO₄) = 331.8 g; mass of solid silver chromate.

n(Ag₂CrO₄) = m(Ag₂CrO₄) ÷ M(Ag₂CrO₄).

n(Ag₂CrO₄) = 331.8 g ÷ 331.8 g/mol.

n(Ag₂CrO₄) = 1 mol; amount of silver chromate.

From balanced chemical reaction: n(Ag₂CrO₄) : n(AgNO₃) = 1 : 2.

n(AgNO₃) = 2 · 1 mol.

n(AgNO₃) = 2 mol.

m(AgNO₃) = n(AgNO₃) · M(AgNO₃).

m(AgNO₃) = 2 mol · 169.8 g/mol.

m(AgNO₃) = 339.6 g; mass of silver nitrate.

m(AgNO₃) = m(K₂CrO₄).

m(K₂CrO₄) = 339.6 g; mass of potassium chromate.

n(K₂CrO₄) = m(K₂CrO₄) ÷ M(K₂CrO₄).

n(K₂CrO₄) = 339.6 g ÷ 194.2 g/mol.

n(K₂CrO₄) = 1.75 mol; amount of potassium chromate.

5) Chemical reaction of dissociation of silver nitrate in water:

AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq).

V(solution a) = 500 mL ÷ 1000 mL/L.

V(solution a) = 0.5 L; volume of solution a.

c(AgNO₃) = n(AgNO₃) ÷ V(solution a).

c(AgNO₃) = 2 mol ÷ 0.5 L.

c(AgNO₃) = 4 mol/L = 4 M.

From dissociation of silver nitrate: c(AgNO₃) = c(Ag⁺) = c(NO₃⁻).

c(Ag⁺) = 4 M; the concentration of silver ions in the original solution a.

c(NO₃⁻) = 4 M; the concentration of silver ions in the original solution a.

6) Chemical reaction of dissociation of potssium chromate in water:

K₂CrO₄(aq) → 2K⁺(aq) + CrO₄²⁻(aq).

V(solution b) = 500 mL ÷ 1000 mL/L.

V(solution b) = 0.5 L; volume of solution b.

c(K₂CrO₄) = n(K₂CrO₄) ÷ V(solution b).

c(AgNO₃) = 1.75 mol ÷ 0.5 L.

c(AgNO₃) = 3.5 mol/L = 3.5 M.

From dissociation of silver nitrate: c(K₂CrO₄) = c/2(K⁺) = c(CrO₄²⁻).

c(K⁺) = 7 M; the concentration of potassium ions in the original solution b.

c(CrO₄²⁻) = 3.5 M; the concentration of silver ions in the original solution b.

7) V(final solution) = V(solution a) + V(solution b).

V(final solution) = 500.0 mL + 500.0 mL.

V(final solution) = 1000 mL ÷ 1000 mL/L.

V(final solution) = 1 L.

n(NO₃⁻) = 2 mol.

c(NO₃⁻) = n(NO₃⁻) ÷ V(final solution)

c(NO₃⁻) = 2 mol ÷ 1 L.

c(NO₃⁻) = 2 M; the concentration of nitrate anions in final solution.

8) in the solution b there were 3.5 mol of potassium cations, but one part of them reacts with 2 moles of nitrate anions:

K⁺(aq) + NO₃⁻(aq) → KNO₃(aq).

From chemical reaction: n(K⁺) : n(NO₃⁻) = 1 : 1.

Δn(K⁺) = 3.5 mol - 2 mol.

Δn(K⁺) = 1.5 mol; amount of potassium anions left in final solution.

c(K⁺) = Δn(K⁺) ÷ V(final solution).

c(K⁺) = 1.5 mol ÷ 1 L.

c(K⁺) = 1.5 M; the concentration of potassium cations in final solution.

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2 years ago

(b) When a sample of C2H5OH was combusted, the volume of CO2(g) produced was 18.0 L when measured at

myrzilka [38]

Answer:

i) The number of moles of CO₂ (g) produced from the reaction = 0.07663 mole

ii) The volume of C₂H₅OH (l), in mL, that was combusted to produce the volume of CO₂ (g)

collected = 2.234 mL

iii) The amount of heat, in KJ, that was released by the combustion reaction = 52.4 kJ

Explanation:

The balanced chemical reaction when ethanol is combusted is given as

C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g)

The volume of CO₂(g) produced was 18.0 L when measured at 21.7°C and 1.03 atm.

i) Number of moles of CO₂ (g) produced by the reaction

With the correct and logical assumption that CO₂ is an ideal gas, the ideal gas equation has the relation

PV = nRT

P = pressure = 1.03 atm = 1.03 × 101325 Pa = 10,435.96 Pa

V = Volume of the gas = 18.0 L = 0.018 m³

n = number of moles = ?

R = molar gas constant = 8.314 J/mol.K

T = absolute temperature in Kelvin = 21.7 + 273.15 = 294.85 K

(10,435.96 × 0.018) = n × 8.314 × 294.85

n = 0.076629106 = 0.07663 mole

ii) The volume of C₂H₅OH (l), in mL, that was combusted to produce the volume of CO₂(g)

collected.

Recall the stoichiometric balance of the reaction

C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g)

2 moles of CO₂ is obtained from 1 mole of C₂H₅OH

0.07663 mole of CO₂ will be obtained from (0.07663×1/2) mole of C₂H₅OH; that is, 0.03831 mole of C₂H₅OH.

But we can convert this number of moles used up to mass of C₂H₅OH produced

Mass = (Number of moles) × (Molar Mass)

Molar mass of C₂H₅OH = 46.07 g/mol

Mass of C₂H₅OH combusted from the reaction

= 0.03831 × 46.07 = 1.765 g

But density of C₂H₅OH = 0.79 g/mL

Density = (Mass)/(Volume)

Volume = (Mass)/(Density) = (1.765/0.79)

= 2.234 mL

iii) The amount of heat, in KJ, that was released by the combustion reaction.

The heat of combustion of C₂H₅OH at the temperature of the reaction = -1367.6 kJ/mol. (From literature)

1 mole of C₂H₅OH combusts to give 1367.6 kJ of heat

0.03831 mole of C₂H₅OH will give (0.03831×1367.6) = 52.39 kJ = 52.4 kJ

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