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].
