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Dominik [7]
3 years ago
10

Defined as the negative logarithm of the concentration of OH-​

Chemistry
1 answer:
Marat540 [252]3 years ago
4 0

Answer:

As with the hydrogen-ion concentration, the concentration of the hydroxide ion can be expressed logarithmically by the pOH. The pOH of a solution is the negative logarithm of the hydroxide-ion concentration. pOH=−log[OH−] The pH of a solution can be related to the pOH.

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Sulfuric acid reacts with aluminum hydroxide by double replacement.
Alinara [238K]

a. <span>
<span>The balanced equation for the reaction between sulfuric acid and aluminium hydroxide is,
<span>                        3H</span>₂SO₄<span>     </span>+   2Al(OH)₃ </span><span>→ Al₂(SO₄)₃<span> +   6H</span>₂O</span><span>

</span></span>

mass of H₂SO₄<span>  =       <span> 30.0 g             </span></span><span>
<span>Molar mass of H</span>₂SO₄</span><span>  =   98 g/mol</span><span>
moles of H₂SO₄</span><span> = 30.0 g /98g /mol = 0.306 mol</span><span>

mass of Al(OH)₃</span><span>            =         25.0 g            </span><span>
Molar mass of Al(OH)₃</span>  =   78 g/mol<span>
moles of Al(OH)₃<span>           </span>= 25.0 g/ 78 g/mol = 0.321 mol</span><span>

Stoichiometric ratio between H₂SO₄<span>  and </span>Al(OH)₃ is 3 : 2</span><span>

Hence reacted moles of H₂SO₄ = 0.306 mol</span><span>
            reacted moles of Al(OH)₃ = 0.306 mol x (2 / 3) = 0.204 mol</span><span>

Hence the limiting reactant is H₂SO₄</span> <span>

b.<span>
</span>According to the above calculation, the excess reactant is Al(OH)₃. </span><span>

The reacted moles of Al(OH)₃<span> = </span>0.306 mol x (2 / 3) = 0.204 mol</span><span>

The added moles of Al(OH)₃ = 0.321 mol</span><span>

Hence the remaining Al(OH)₃ moles = added moles - reacted moles</span><span>
                                                          = 0.321 mol - 0.204 mol
                                                          = 0.117 mol

Molar mass of Al(OH)₃</span>  =   78 g/mol<span>
<span>Remaining mass of Al(OH)</span>₃ = number of moles x molar mass</span><span>
                                             = 0.117 mol x 78 g/mol
                                             = 9.126 g
</span><span><span>
</span>c. 
</span>

The products formed from the reaction between aluminium hydroxide and sulfuric acid are Al₂(SO₄)₃<span> and </span>H₂O<span>

The limiting reactant is H₂SO₄</span> <span>

The stoichiometric ratio between H₂SO₄  and Al₂(SO₄)₃</span> is 3 : 1<span>
Reacted moles of H₂SO₄ = 0.306 mol</span><span>
Hence the moles of Al₂(SO₄)₃ formed = 0.306 mol / 3</span><span>
                                                             = 0.102 mol
Molar mass of Al₂(SO₄)₃</span>  = 342 g/mol<span>
Mass of Al₂(SO₄)₃</span>  formed = 0.102 mol x 342 g/mol<span>
<span>                                           = </span>34.884 g

The stoichiometric ratio between H₂SO₄  and H₂O is 3 : 6</span><span>
Reacted moles of H₂SO₄ = 0.306 mol</span><span>
Hence the moles of H₂O formed = 0.306 mol x (6 / 3)</span><span>
                                                    = 0.612 mol
Molar mass of H₂O  = 18 g/mol</span><span>
Mass of H₂O  formed = 0.612 mol x 18 g/mol</span><span>
<span>                                   = </span><span>11.016 g</span></span>


5 0
3 years ago
How many molecules are there in 5.0 g of methyl alcohol, ch3oh?
Step2247 [10]
5.0 g CH3OH x 1 mol
32.05 g
x 6.02x10^23 molecules = answer
1 mol
6 0
3 years ago
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