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Fall 2109 CHE 151-01 Exam 2 Chapter 4-6 Question 5 Question 6 (21 Points) Al2S3 + H2O - Al(OH)3 + H2S Suppose 316.0 g aluminum sulfide reacts with 0.493 kg of water. a) Write the balanced equation (2 Points) b) What % of Al(OH)3 is Al by mass? (2 Points) c) Identify the limiting and excess reagents (2Points) d) Calculate the number of moles and the mass in (g) after the reaction is complete for: a. The excess reagent (3 Points) b. Al(OH)3 (3 Points) c. H2S (3 Points) d. How many molecules of hydrogen sulfide were produced (3 Points) e. If 27 g of aluminum are actually obtained by a student in the experiment, what was the %yield of the student's reaction (3 Points)?

Answer 1

a) The balanced equation will be

$Al_2S_3+6H_2O\rightarrow 2Al(OH)_3+3H_2S$

b)

Molar Mass of Al(OH)₃ = 27+ 3 * (16+1) g/mol =78 g/mol

Mass of Aluminum in 1 mol Al(OH)₃ = 27 g

% of Aluminum = Mass of Aluminum/ Mass of Al(OH)₃ * 100 =27/78*100 =34.62 %

c)

Molar Mass of Al₂S₃ = 150.158 g/mol

Moles of Al₂S₃= Mass/Molar Mass = 316.0/150.158= 2.104 moles

Molar Mass of H₂O = 18.0 g/mol

Moles of H₂O = Mass/Molar Mass =493/18= 27.39 moles

1 mole of Al₂S₃ needs 6 moles water. Thus, 2.104 moles Al₂S₃ needs 2.104 *3 = 12.624 moles water. But water is 27.39 moles.

Thus, limiting reagent is Al₂S₃ and excess reagent is water.

d)

i)

Moles of water reacted = 12.624moles

Mass of water reacted = 12.624 * 18 g = 227.232 g

Mass of Water left = Total Mass - Mass reacted = 493 g- 227.232 g= 265.768 g = 266 g (approx)

Moles of Water left = Mass left/ Molar Mass = 227.232/18 mol = 14.76 mol = 14.8 mol (approx)

ii)

Moles of Al(OH)₃ = 2 * Moles of Al₂S = 2* 2.104 mol = 4.208 mol = 4.21 mol (approx)

Mass of Al(OH)₃= Moles * Molar Mass = 4.208 * 150.158 g = 631.9 g (approx)

iii)

Moles of H₂S = 3 * Moles of Al₂S = 3* 2.104 mol = 6.312 mol = 6.31 mol (approx)

Molar Mass of H₂S = 34.1 g/mol

Mass of H₂S= Moles * Molar Mass = 6.312 * 34.1 g = 215 .2 g (approx)

iv)

Molecules of H₂S = Moles * Avogadro's Constant = 6.31 * 6.022 *10²³= 3.80*10²⁴ molecules

v)

27 g Al = 1 mol Aluminum

1 mol Aluminum is present in 1 mol Al(OH)₃

Thus, 1 mol Al(OH)₃ was formed. But expected Yield was 4.21 mol

Thus % Yield = Actual Yield/Expected Yield * 100 = 1/4.21*100 = 23.75 %