Respuesta :
Answer:
6.935g
Explanation:
From the question above, we can see that 1 mole of magnesium hydroxide neutralized 2 moles of hydrochloric acid.
Now, let's calculate the actual number of moles of magnesium hydroxide reacted. We can do this by dividing the mass of the magnesium hydroxide by the molar mass. Molar mass of the magnesium hydroxide is 24 + 2(17) = 58g/mol
The number of moles thus produced is 5.5/58 = 0.095moles
From the first relation we established that 1 mole of magnesium hydroxide reacted with 2 moles of hydrochloric acid. Hence, 0.095moles of magnesium hydroxide will yield 2 × 0.095 moles of hydrochloric acid = 0.190 moles
We then calculate the mass of HCl that be neutralized by multiplying the number of moles by the molar mass. The molar mass of HCl is 1 + 35.5 = 36.5g/mol.
The mass of HCl neutralized = 36.5 × 0.190 = 6.935g
Answer: The mass of [tex]HCl[/tex] neutralized can be, 6.88 grams.
Explanation : Given,
Mass of [tex]Mg(OH)_2[/tex] = 5.50 g
Molar mass of [tex]Mg(OH)_2[/tex] = 58.3 g/mol
Molar mass of [tex]HCl[/tex] = 36.5 g/mol
First we have to calculate the moles of [tex]Mg(OH)_2[/tex]
[tex]\text{Moles of }Mg(OH)_2=\frac{\text{Given mass }Mg(OH)_2}{\text{Molar mass }Mg(OH)_2}[/tex]
[tex]\text{Moles of }Mg(OH)_2=\frac{5.50g}{58.3g/mol}=0.0943mol[/tex]
Now we have to calculate the moles of [tex]HCl[/tex]
The balanced chemical equation is:
[tex]Mg(OH)_2(aq)+2HCl(aq)\rightarrow 2H_2O(l)+MgCl_2(aq)[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]Mg(OH)_2[/tex] react with 2 mole of [tex]HCl[/tex]
So, 0.0943 moles of [tex]Mg(OH)_2[/tex] react with [tex]0.0943\times 2=0.1886[/tex] moles of [tex]HCl[/tex]
Now we have to calculate the mass of [tex]HCl[/tex]
[tex]\text{ Mass of }HCl=\text{ Moles of }HCl\times \text{ Molar mass of }HCl[/tex]
[tex]\text{ Mass of }HCl=(0.1886moles)\times (36.5g/mole)=6.88g[/tex]
Therefore, the mass of [tex]HCl[/tex] neutralized can be, 6.88 grams.