How many moles of magnesium chloride are formed when [tex]$1.204 \, g \, Mg(OH)_2$[/tex] is added to 55 mL of 0.70 M HCl?

[tex]\[ Mg(OH)_2(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + 2H_2O(l) \][/tex]

A. 0.039 mol
B. 0.019 mol
C. 0.021 mol
D. 0.012 mol



Answer :

To determine how many moles of magnesium chloride ([tex]\( \text{MgCl}_2 \)[/tex]) are formed when 1.204 grams of magnesium hydroxide ([tex]\( \text{Mg(OH)}_2 \)[/tex]) are added to 55 mL of 0.70 M hydrochloric acid ([tex]\( \text{HCl} \)[/tex]), we need to follow several steps:

1. Convert given masses and volumes to moles:

- Step 1: Calculate moles of [tex]\( \text{Mg(OH)}_2 \)[/tex]:
- Molar mass of [tex]\( \text{Mg(OH)}_2 \)[/tex]:
[tex]\[ \text{Mg:} 24.305 \, \text{g/mol} \\ \text{O:} 15.999 \, \text{g/mol} \, \text{(each)} \\ \text{H:} 1.008 \, \text{g/mol} \, \text{(each)} \\ \text{Total molar mass of} \, \text{Mg(OH)}_2: 24.305 + 2(15.999 + 1.008) = 58.3197 \, \text{g/mol} \][/tex]
- Moles of [tex]\( \text{Mg(OH)}_2 \)[/tex]:
[tex]\[ \frac{1.204 \, \text{g}}{58.3197 \, \text{g/mol}} = 0.020644824990526358 \, \text{mol} \][/tex]

- Step 2: Calculate moles of [tex]\( \text{HCl} \)[/tex]:
- Molarity (M) = moles/volume (L)
- Volume of [tex]\( \text{HCl} \)[/tex]:
[tex]\[ 55 \, \text{mL} = 0.055 \, \text{L} \][/tex]
- Moles of [tex]\( \text{HCl} \)[/tex]:
[tex]\[ 0.70 \, \text{M} \times 0.055 \, \text{L} = 0.0385 \, \text{mol} \][/tex]

2. Determine the limiting reactant:

According to the balanced chemical equation:
[tex]\[ \text{Mg(OH)}_2(s) + 2 \text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + 2 \text{H}_2\text{O}(l) \][/tex]
- One mole of [tex]\( \text{Mg(OH)}_2 \)[/tex] reacts with two moles of [tex]\( \text{HCl} \)[/tex].

- Calculate the amount of [tex]\( \text{HCl} \)[/tex] required for the given moles of [tex]\( \text{Mg(OH)}_2 \)[/tex]:
[tex]\[ 0.020644824990526358 \, \text{mol} \, \text{Mg(OH)}_2 \times 2 = 0.041289649981052716 \, \text{mol} \, \text{HCl} \][/tex]

- Compare this with the available moles of [tex]\( \text{HCl} \)[/tex]:
[tex]\[ \text{Available moles of } \text{HCl} = 0.0385 \, \text{mol} \][/tex]
- Since [tex]\( 0.0385 \, \text{mol HCl} < 0.041289649981052716 \, \text{mol HCl} \)[/tex], [tex]\( \text{HCl} \)[/tex] is the limiting reactant.

3. Calculate moles of [tex]\( \text{MgCl}_2 \)[/tex] formed:

- Based on the stoichiometry of the reaction, 2 moles of [tex]\( \text{HCl} \)[/tex] produce 1 mole of [tex]\( \text{MgCl}_2 \)[/tex].
- Moles of [tex]\( \text{MgCl}_2 \)[/tex] formed:
[tex]\[ \frac{0.0385 \, \text{mol} \, \text{HCl}}{2} = 0.01925 \, \text{mol} \][/tex]

Thus, the number of moles of magnesium chloride ([tex]\( \text{MgCl}_2 \)[/tex]) formed is [tex]\( 0.019 \)[/tex] mol, which corresponds to option B.

So, the correct answer is [tex]\( \mathbf{B. \, 0.019 \, mol} \)[/tex].