Question

Solve the given problem

 3 HCI + AI(OH)3 --> 3 H2O AICI3

1) How many grams og aluminum hydroxide will react 5.3 moles of HCI?

 2 NaOH + H2SO4 --> 2 H20 + Na2SO4

2) How many grams of solution of sofium sulfatr will be formed, if you start with 1.25 L of a 4.0 M solution of sodium hydroxide?

Answer 1

Q181304

Solve the given problem

Question 1

 3 HCI + AI(OH)₃ --> 3 H₂O + AICI₃

How many grams of aluminum hydroxide will react 5.3 moles of HCI?

 Solution :

In the given reaction the mol to mol ratio of HCl and Al(OH)₃ is 3 :1 .

Using this ratio find the moles of Al(OH)₃ that will react with 5.3 moles of HCl .

$$mol \ of \ AlCl_3 = 5.3 \ mol \ HCl * \frac{1 \ mol \ Al(OH)_3}{3 \ mol \ HCl } = 1.77 \ mol \ Al(OH)_3$$

Next convert 1.77 mol Al(OH)₃ to grams by using molar mass of Al(OH)₃.

molar mass of Al(OH)₃ = 1 * atomic mass of Al + 3 * atomic mass of O + 3 * atomic mass of H.

= 1 * 26.982 g/mol + 3 * 15.999 g/mol + 3 * 1.00794 g/mol

= 78.0028 g/mol

Using molar mass 78.0028 g/mol convert 1.77 mol Al(OH)₃ to grams.

$$grams \ of \ Al(OH)_3 = 1.77 \ mol \ Al(OH)3 \ * \frac{78.0028 \ g \ Al(OH)_3 }{1 \ mol \ Al(OH)_3 }$$

$$= 138.06\ grams \ of \ Al(OH)_3$$

In question we are given 5.3 mole HCl in 2 significant figure, so our final answer must also be in

2 significant figure.

Hence the mass of Al(OH)₃ that will react with 5.3 mole HCl is 140 grams.

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Question 2

  2 NaOH + H₂SO₄ --> 2 H₂0 + Na₂SO₄

How many grams of solution of sodium sulfate will be formed, if you start with 1.25 L of a 4.0 M solution of sodium hydroxide?

Solution :

Step 1 : To find the moles of NaOH present in 1.25 L of 4.0 M solution.

Let us first find the moles of NaOH present in 1.25 L of 4.0 M solution of sodium hydroxide.

Plug volume = 1.25 L and molarity = 4.0 M in the molarity formula and find the

moles of NaOH.

$$Molarity = \frac{moles \ of \ NaOH }{volume \ of \ solution \ in \ 'L' }$$

$$4.0M = \frac{moles \ of \ NaOH }{1.25 \ L }$$

Arranging the equation for ' moles of NaOH' , we have

$$moles \ of \ NaOH \ = \ 4.0 \ mol/L * {1.25 \ L } \ = \ 5.0 \ moles \ of \ NaOH.$$

Step 2 : To find the moles of Na₂SO₄ that is formed from 5.0 moles of NaOH.

The reaction of NaOH and H₂SO₄ is

2 NaOH + H₂SO₄ --> 2 H₂0 + 1 Na₂SO₄

The mol to mol ratio of NaOH and Na₂SO₄ in this reaction is 2 mol NaOH : 1 mol Na₂SO₄.

Using this ratio find the moles of Na₂SO₄ that will be formed from 5.0 moles of NaOH.

$$moles \ of \ Na_2SO_4 = 5.0 \ mol \ of \ NaOH * \frac{1 \ mol \ Na_2SO_4}{2 \ mol \ NaOH }$$

$$= 2.5 \ mol \ Na_2SO_4$$

Step 3 : Convert 2.5 mol Na₂SO₄ to grams by using molar mass of Na₂SO₄.

molar mass of Na₂SO₄ = 2 * atomic mass of Na + 1 * atomic mass of S + 4 * atomic mass of O

= 2 * 22.99 g/mol + 1 * 32.065 g/mol + 4 * 15.999 g/mol

= 142.041 g/mol

Using molar mass 142.041 g/mol convert 2.5 moles of Na₂SO₄ to grams.

$$grams of Na_2SO_4 = 2.5 \ mol \ Na_2SO_4 * \frac{142.041 \ g \ of \ Na_2SO_4 }{1 \ mol \ of \ Na_2SO_4}$$

$$= 355.1 \ grams \ of \ Na_2SO_4$$

In correct significant figure the answer will be 360 grams of Na₂SO₄.

The molarity given in the question is in 2 significant figure ( which is least)

So our final answer must also be in 2 significant figure.

Hence 360 grams of Na₂SO₄ will be formed from the given 1.25 L of a 4.0 M solution of sodium hydroxide