To determine how many moles of [tex]\( \text{BaCl}_2 \)[/tex] form from 100.0 mL of 1.0 M HCl when reacted with excess [tex]\( \text{Ba(OH)}_2 \)[/tex], follow these steps:
1. Convert the volume of HCl from mL to L:
We are given a volume of 100.0 mL. To convert this to liters, recall that:
[tex]\[
1 \, \text{L} = 1000 \, \text{mL}
\][/tex]
Therefore,
[tex]\[
\text{Volume of HCl} = \frac{100.0 \, \text{mL}}{1000} = 0.1 \, \text{L}
\][/tex]
2. Calculate the moles of HCl:
We are given the concentration of HCl as 1.0 M (moles per liter). The formula to calculate moles from volume and concentration is:
[tex]\[
\text{Moles of HCl} = \text{Concentration of HCl} \times \text{Volume in liters}
\][/tex]
Substituting the values, we get:
[tex]\[
\text{Moles of HCl} = 1.0 \, \text{M} \times 0.1 \, \text{L} = 0.1 \, \text{moles}
\][/tex]
3. Determine the moles of [tex]\( \text{BaCl}_2 \)[/tex] produced:
The balanced chemical equation for the reaction is:
[tex]\[
\text{Ba(OH)}_2 + 2 \, \text{HCl} \rightarrow \text{BaCl}_2 + 2 \, \text{H}_2\text{O}
\][/tex]
From the stoichiometry of the reaction, 2 moles of HCl react with 1 mole of [tex]\( \text{Ba(OH)}_2 \)[/tex] to produce 1 mole of [tex]\( \text{BaCl}_2 \)[/tex].
Therefore, for every 2 moles of HCl, we get 1 mole of [tex]\( \text{BaCl}_2 \)[/tex]. We have 0.1 moles of HCl, so we can find the moles of [tex]\( \text{BaCl}_2 \)[/tex] produced by dividing by 2:
[tex]\[
\text{Moles of BaCl}_2 = \frac{0.1 \, \text{moles of HCl}}{2} = 0.05 \, \text{moles}
\][/tex]
Hence, from 100.0 mL of 1.0 M HCl and excess [tex]\( \text{Ba(OH)}_2 \)[/tex], we form 0.05 moles of [tex]\( \text{BaCl}_2 \)[/tex].
