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2 years ago

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

Chemistry

2 answers:

almond37 [142]2 years ago

7 0

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 = $6.72 \times 10^{-2} mol$

Thus, we can conclude that the number of moles of $BaCl_{2}$ formed are $6.72 \times 10^{-2} mol$ .

Scilla [17]2 years ago

4 0

Answer:

$n_{BaCl_2}=0.0672molBaCl_2$

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:

$n_{BaCl_2}=0.171\frac{molBa(OH)_2}{L}*0.393L*\frac{1mol BaCl_2}{1mol Ba(OH)_2} \\n_{BaCl_2}=0.0672molBaCl_2$

Best regards.

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