Question

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?

Answer 1

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

`\tt (35)/(115.854)=0.302`

• mol Fe₂O₃

`\tt (2)/(4)* 0.302=0.151`

• mass Fe₂O₃

`\tt 0.151* 159.69=24.113~g`

• percent yield

`\tt (22.5)/(24.113)* 100\%=93.3\%`

mol of 1 kg Fe₂O₃ = 1000 g

`\tt (1000)/(159.69)=6.262`

mol of FeCO₃

`\tt (4)/(2)* 6.262=12.524`

mass of FeCO₃

`\tt 12.524* 115.854=1450.955~g`

a purity of 62.8%

`\tt (100)/(62.8) * 1450.955=2310.44~g`