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True [87]
1 year ago
11

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.​
Chemistry
1 answer:
alex41 [277]1 year ago
8 0

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

<h3>What is neutralization?</h3>

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: brainly.com/question/202460

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1(0.200M)(Vol Am-OH Soln) = 3(0.550M)(0.012L)

=> Vol Am-OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liter = 9.9 milliliters  

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What mass of C6H8O7 should be used every 7.0 X 10^2mg NaHCO3
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Mass C₆H₈O₇ : 0.531484 g

<h3>Further explanation</h3>

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3NaHCO₃ (aq) + C₆H₈O₇ (aq) → 3 CO₂ (g) + 3 H₂O (l) + Na₃C₆H₅O₇ (aq)

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