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4 FeCO3 + O2 --> 2 Fe2O3 + 4CO2

a) A 35.0 g sample of pure FeCO3 produces 22.5 g of Fe2O3. What is the percentage yield of the reaction?

b) What mass of FeCO3 with a purity of 62.8% is needed to make 1.00 kg of Fe2O3?

Respuesta :

The percentage yield of the reaction : 93.3%

Mass of FeCO₃ 2310.44 g

Further explanation

The reaction equation is the chemical formula of reagents and product substances  

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products  

Reaction

4 FeCO₃ + O₂ ⇒ 2 Fe₂O₃ + 4CO₂

MW FeCO₃ : 115,854 g/mol

MW Fe₂O₃ : 159,69 g/mol

  • mol FeCO₃

[tex]\tt \dfrac{35}{115.854}=0.302[/tex]

  • mol Fe₂O₃

[tex]\tt \dfrac{2}{4}\times 0.302=0.151[/tex]

  • mass Fe₂O₃

[tex]\tt 0.151\times 159.69=24.113~g[/tex]

  • percent yield

[tex]\tt \dfrac{22.5}{24.113}\times 100\%=93.3\%[/tex]

mol of 1 kg Fe₂O₃ = 1000 g

[tex]\tt \dfrac{1000}{159.69}=6.262[/tex]

mol of FeCO₃

[tex]\tt \dfrac{4}{2}\times 6.262=12.524[/tex]

mass of FeCO₃

[tex]\tt 12.524\times 115.854=1450.955~g[/tex]

a purity of 62.8%

[tex]\tt \dfrac{100}{62.8} \times 1450.955=2310.44~g[/tex]