Explanation:
As there are 1000 mL present in 1 L. So, 393 ml will be equal to 0.393 L. And, concentration of [tex]Ba(OH)_{2}[/tex] present is 0.171 M.
As molarity is the number of moles present in liter of solution.
Molarity = [tex]\frac{\text{no. of moles}}{\text{Volume in liter}}[/tex]
Therefore, putting the given values to calculate the number of moles as follows.
Molarity = [tex]\frac{\text{no. of moles}}{\text{Volume in liter}}[/tex]
0.171 M = [tex]\frac{\text{no. of moles}}{0.393 L}[/tex]
no. of moles = [tex]6.72 \times 10^{-2} mol[/tex]
Thus, we can conclude that the number of moles of [tex]BaCl_{2}[/tex] formed are [tex]6.72 \times 10^{-2} mol[/tex].
Answer:
[tex]n_{BaCl_2}=0.0672molBaCl_2[/tex]
Explanation:
Hello,
In this case, the stoichiometry is given by the undergoing chemical reaction which is:
[tex]2HCl+Ba(OH)_2-->BaCl_2+2H_2O[/tex]
Now, as the neutralization is carried out to completion, the resulting moles of barium chloride are given by the following stoichiometric relationship including the molarity unit, considering the 1 to 1 relationship between barium hydroxide and barium chloride and that the volume is used in liter rather than in milliliters:
[tex]n_{BaCl_2}=0.171\frac{molBa(OH)_2}{L}*0.393L*\frac{1mol BaCl_2}{1mol Ba(OH)_2} \\n_{BaCl_2}=0.0672molBaCl_2[/tex]
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