Question

17. Consider the following balanced chemical equation:

A. 2Al(s) + 3Cl₂(g) → 2AlCl₃(s)   Determine the mass (in g) of AlCl₃ formed if 27.6 g of Al reacts with 47.6 g of Cl₂

B. What volume, in L, of 0.0771 M LiOH solution is required to produce 72.9 g of water.

2LiOH(aq) + H₂SO₄(aq) → 2H₂O(l) + Li₂SO₄(aq)

_C. What volume (in mL, at 558 K and 3.49 atm) of oxygen gas is required to react with 4.84 g of Al?

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

Answer 1

A)

Molar mass of Al = 26.98 g/mol

mass(Al)= 27.6 g

use:

number of mol of Al,

n = mass of Al/molar mass of Al

=(27.6 g)/(26.98 g/mol)

= 1.023 mol

Molar mass of Cl2 = 70.9 g/mol

mass(Cl2)= 47.6 g

use:

number of mol of Cl2,

n = mass of Cl2/molar mass of Cl2

=(47.6 g)/(70.9 g/mol)

= 0.6714 mol

Balanced chemical equation is:

2 Al + 3 Cl2 ---> 2 AlCl3

2 mol of Al reacts with 3 mol of Cl2

for 1.023 mol of Al, 1.534 mol of Cl2 is required

But we have 0.6714 mol of Cl2

so, Cl2 is limiting reagent

we will use Cl2 in further calculation

Molar mass of AlCl3,

MM = 1*MM(Al) + 3*MM(Cl)

= 1*26.98 + 3*35.45

= 133.33 g/mol

According to balanced equation

mol of AlCl3 formed = (2/3)* moles of Cl2

= (2/3)*0.6714

= 0.4476 mol

use:

mass of AlCl3 = number of mol * molar mass

= 0.4476*1.333*10^2

= 59.68 g

Answer: 59.7 g

B)

Balanced chemical equation is:

2LiOH(aq) + H2SO4(aq) → 2H2O(l) + Li2SO4(aq)

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 72.9 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(72.9 g)/(18.02 g/mol)

= 4.046 mol

According to balanced equation

mol of LiOH reacted = (2/2)* moles of H2O

= (2/2)*4.046

= 4.046 mol

This is number of moles of LiOH

use:

M = number of mol / volume in L

0.771 = 4.046/ volume in L

volume = 5.25 L

Answer: 5.25 L

C)

Molar mass of Al = 26.98 g/mol

mass of Al = 4.84 g

mol of Al = (mass)/(molar mass)

= 4.84/26.98

= 0.1794 mol

According to balanced equation

mol of O2 required = (3/4)* moles of Al

= (3/4)*0.1794

= 0.1345 mol

Given:

P = 3.49 atm

n = 0.1345 mol

T = 558 K

use:

P * V = n*R*T

3.49 atm * V = 0.1345 mol* 0.08206 atm.L/mol.K * 558 K

V = 1.76 L

Answer: 1.76 L