A sample of gasoline contains various hydrocarbons, which comprise atoms of carbon, hydrogen, and oxygen. The hydrocarbons mix t

Answer:

Explanation:

This scenario is unrealistic since Al(OH)₃ is not soluble in water.

The question consists of two parts:

A. Stoichiometry — where we determine volumes, masses, and moles for the products

B. Calorimetry — where we assess the enthalpy of the reaction.

A. Stoichiometry

1. Determine the volume of Al(OH)₃

(a) The balanced chemical equation:

                 2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 6H₂O

M/V:            66.667

c/mol·L⁻¹:   4.000       3.000

(b) Moles of H₂SO₄

(c) Moles of Al(OH)₃

The molar ratio stands at 2 mmol Al(OH)₃: 3 mmol H₂SO₄

(d) Volume of Al(OH)₃

B. Calorimetry

This reaction has two energy exchanges.

q₁ = heat from the reaction

q₂ = heat used to heat the calorimeter

 q₁ + q₂ = 0

nΔH + mCΔT = 0

Data:

Moles of Al₂(SO₄)₃ = 0.066 667 mol

C = 1.10 J°C⁻¹g⁻¹

T_initial = 22.3 °C

T_final = 24.7 °C

Calculations

(a) Mass of solution

Assume solutions are as dense as water (though not realistic).

Mass of sulfuric acid solution            =   66.667 g 

Mass of aluminium hydroxide solution =  50.000    

                                             TOTAL =  116.667 g

(b) ΔT

ΔT = T_final - T_initial = 24.7 °C - 22.3 °C = 2.4°C

(c) ΔH

This result appears nonsensical, but it is derived from your given figures.

1) Calcium carbonate comprises 40.0% calcium by weight.
M(CaCO₃)=100.1 g/mol
M(Ca)=40.1 g/mol
w(Ca)=40.1/100.1=0.400 (which is 40.0%)!

2) The mass fraction mentioned is superfluous information.

3) The resulting solution is:

m(Ca)=1.2 g

m(CaCO₃)=M(CaCO₃)*m(Ca)/M(Ca)

m(CaCO₃)=100.1g/mol*1.2g/40.1g/mol=3.0 g

Start by determining the number of moles, which is obtained by dividing 38 grams by the molar mass of 58.43 g/mol. This calculation yields 0.65 moles. The concentration is calculated by dividing the number of moles by the volume in liters. Using this formula, we can derive the total volume by dividing the number of moles by the concentration. Thus, 0.65 moles divided by 0.02M (mol/L) results in a total volume of 32.5 L.

<span>128 g/mol Applying Graham's law of effusion, we can utilize the formula: r1/r2 = sqrt(m2/m1) where r1 = effusion rate of gas 1 r2 = effusion rate of gas 2 m1 = molar mass of gas 1 m2 = molar mass of gas 2 Given that the atomic weight of oxygen is 15.999, the molar mass of O2 = 2 * 15.999 = 31.998. We can now insert the known values into Graham's equation to find m2. r1/r2 = sqrt(m2/m1) 2/1 = sqrt(m2/31.998) 4/1 = m2/31.998 Thus, we find m2 to be 127.992. Rounding to three significant figures yields 128 g/mol</span>