Chemistry: A 0.573 g portion of an ammonium salt containing a single mole of ammonium ion for each mole of the salt was reacted with 50.00 mL of 0.2000 M NaOH. After the reaction the solution was boiled to remove ammonia created during the reaction. The resulting solution was back titrated to the end point with 26.80 mL of 0.1000 M H_2C_2O_4. 1. Calculate the number of moles of hydroxide ion in the 50.00 mL of 0.2000 M sodium hydroxide solution added to the ammonium salt. 2. From the volume and concentration

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A 0.573 g portion of an ammonium salt containing a

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22.01.2022

1. 0.01000 moles of NaOH

2. 0.002680 moles of H₂C₂O₄

3. 0.005360 moles of NaOH

4. 0.004640 moles of NaOH

5. 0.004640 moles of ammonium salt

6. 123.5 g/mol

Explanation:

Ammonium ion (NH₄⁺) reacts with NaOH, thus:

NH₄⁺ + NaOH → NH₃ + H₂O + Na⁺ (1)

1. 50.00mL of 0.2000M NaOH contains:

0.05000L × (0.2000 mol / 1L) = 0.01000 moles of NaOH

2. Oxalic acid (H₂C₂O₄) reacts with NaOH, moles of oxalic acid used during titration are:

0.02680L × (0.1000 mol / 1L) = 0.002680 moles of H₂C₂O₄

3. The reaction of oxalic acid with NaOH is:

H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + H₂O (2)

That means one mole of acid reacts with 2 moles of NaOH. Thus, moles of NaOH that remains after reaction with ammonium salt are:

0.002680 moles of H₂C₂O₄ × (2 mol NaOH / 1 mol H₂C₂O₄) = 0.005360 moles of NaOH

4. Thus, moles of NaOH that reacted with ammonium salt are:

0.01000 moles of NaOH - 0.005360 moles of NaOH = 0.004640 moles of NaOH

5. Based in reaction (1), moles of NaOH ≡ moles of NH₄⁺. That means moles of ammonium salt in 0.573g of sample are 0.004640 moles of ammonium salt

6. As moles of ammonium ion are the same than ammonium salt. Molar mass of the ammonium salt is:

0.573g / 0.004640 moles = 123.5 g/mol

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