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21.10g of NaOH and Ba3(OH)2 mixture is dissolved water to prepare 1.0dm³ Solution. To neutralize 25.OO mL of this solution needs 0.5 moldm-³ HCl 15.00mL. calculate the percentage of NaOH by mass in the mixture.​

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From the equation of te reaction, we know that the mass percent  of NaOH in the mixture is 1.4%.

What is neutralization?

Neutralization is a reaction that occurs between an acid and a base to yield salt and water only.

In tis case, the reaction of the NaOH and HCl occurs as follows; NaOH + HCl ----> NaCl + H2O

Number of moles of HCl reacted = 15/1000 * 0.5 moldm-³ = 0.0075 moles

Since the reaction is 1:1, 0.0075 moles of NaOH reacted.

Mass of NaOH =  0.0075 moles of NaOH * 40 g/mol = 0.3 g

Percent of NaOH = 0.3 g/21.10g * 100/1 = 1.4%

Learn more about percent concentration: https://brainly.com/question/202460

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