Question

How many moles of BaCl2 are formed in the neutralization of 393 mL of 0.171 M Ba(OH)2 with aqueous HCI

Answer 1

`n_(BaCl_2)=0.0672molBaCl_2`

Step-by-step explanation:

Hello,

In this case, the stoichiometry is given by the undergoing chemical reaction which is:

`2HCl+Ba(OH)_2-->BaCl_2+2H_2O`

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:

`n_(BaCl_2)=0.171(molBa(OH)_2)/(L)*0.393L*(1mol BaCl_2)/(1mol Ba(OH)_2)  \\n_(BaCl_2)=0.0672molBaCl_2`

Best regards.

Answer 2

Step-by-step explanation:

As there are 1000 mL present in 1 L. So, 393 ml will be equal to 0.393 L. And, concentration of
`Ba(OH)_(2)` present is 0.171 M.

As molarity is the number of moles present in liter of solution.

Molarity =
`\frac{\text{no. of moles}}{\text{Volume in liter}}`

Therefore, putting the given values to calculate the number of moles as follows.

Molarity =
`\frac{\text{no. of moles}}{\text{Volume in liter}}`

0.171 M =
`\frac{\text{no. of moles}}{0.393 L}`

no. of moles =
`6.72 * 10^(-2) mol`

Thus, we can conclude that the number of moles of
`BaCl_(2)` formed are
`6.72 * 10^(-2) mol`.