Does anyone know how to do question nine if can you please show working out. thanks.

A. 257600g (i.e 257.6 Kg) of CaO were produced.

B. 202400g (i.e 202.4 Kg) of CO2 were produced.

Explanation:

We'll begin by calculating the mass of CaCO3 present in 5×10² Kg of lime stone. This is illustrated below:

From the question given, the lime stone contains 92% of CaCO3.

Mass of CaCO3 = 92% x 5×10² Kg

Mass of CaCO3 = 92/100 x 5×10² Kg

Mass of CaCO3 = 460 Kg

Therefore, 460 kg of CaCO3 is present in the lime stone.

Next, we shall determine the mass of CaCO3 that was heated and the masses of CaO and CO2 produced from the balanced equation. This is illustrated below:

CaCO3(s) —> CaO(s) + CO2(g)

Molar mass of CaCO3 = 40 + 12 + (16x3) = 100g/mol

Mass of CaCO3 from the balanced equation = 1 x 100 = 100g

Molar mass of CaO = 40 + 16 = 56g

Mass of CaO from the balanced equation = 1 x 56 = 56g

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g

Summary:

From the balanced equation above,

100g of CaCO3 were heated to produce 56g of CaO and 44g of CO2.

A. Determination of the mass of quick lime, CaO produced by heating 5×10² Kg of lime stone.

5×10² Kg of lime stone contains 460 Kg (i.e 460×10³ g) of CaCO3.

From the balanced equation above,

100g of CaCO3 were heated to produce 56g of CaO.

Therefore, 460×10³g of CaCO3 will be heated to produce = (460×10³ x 56)/100 = 257600g of CaO.

Therefore, 257600g (i.e 257.6 Kg) of CaO were produced.

B. Determination of mass of carbon dioxide, CO2 produced by heating 5×10² Kg of lime stone.

5×10² Kg of lime stone contains 460 Kg (i.e 460×10³ g) of CaCO3.

From the balanced equation above,

100g of CaCO3 were heated to produce 44g of CO2.

Therefore, Therefore, 460×10³g of CaCO3 will be heated to produce = (460×10³ x 44)/100 = 202400g of CO2.

Therefore, 202400g (i.e 202.4 Kg) of CO2 produced.

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