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nekit [7.7K]
3 years ago
11

Lithium arsenate (Li3AsO4) and iron nitrate (Fe(NO3)3) are dissolved in water. What precipitate would you expect to form? what i

s the name and formula?
Chemistry
1 answer:
Alla [95]3 years ago
8 0

Answer:

  • CO₃²⁻ +  H₂O <---------------->  HCO₃⁻ + OH⁻
  • The chemical formula of the precipitate is Fe(OH)₃

Explanation:

  • Fe(NO₃)₃ and K₂CO₃ are strong electrolytes and completely dissociate in water. Carbonate ions is a weak base and combine with water to form hydroxide ions (OH⁻), CO₃²⁻ +  H₂O <---------------->  HCO₃⁻ + OH⁻
  • Ferric, Fe (III), combines with these hydroxide ions to form insoluble precipitates. Fe(OH)₃ is only partially soluble i.e., it does not completely dissociate in water. When the solutions of Fe(NO₃)₃ and K₂CO₃ are mixed, Fe(OH)₃ precipitates out due to the strong electrostatic attraction between Fe (III) and hydroxide ions.
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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa
notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4

Finally, the percent yield turns out into:

Y=\frac{1.92g}{7.1g} *100

Y=27.0%

Best regards.

6 0
3 years ago
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