Calculating Mass of Limestone for Cement Production

How can we calculate the mass of limestone required to make 25 kg of cement?

Answer:

To calculate the mass of limestone required to make 25 kg of cement, use stoichiometry and the molar mass of calcium carbonate.

Final answer: To calculate the mass of limestone required to make 25 kg of cement, use stoichiometry and the molar mass of calcium carbonate. 4 moles of CaCO3 are required to produce 2 moles of Ca3SiO5, the main component of cement. From there, you can calculate the mass of limestone needed.

Explanation: To calculate the mass of limestone required to make 25 kg of cement, we need to use stoichiometry and the molar mass of calcium carbonate (CaCO3). From the balanced equation, we can see that 4 moles of CaCO3 are required to produce 2 moles of Ca3SiO5, which is the main component of cement.

First, we need to calculate the number of moles of Ca3SiO5 in 25 kg of cement using the molar mass provided:

  • Mass of cement = 25 kg
  • Molar mass of cement = 446.6 g/mol
  • Number of moles of Ca3SiO5 = (Mass of cement) / (Molar mass of cement)

Next, using the stoichiometry of the reaction, we can determine the number of moles of CaCO3 required:

  • According to the balanced equation, 4 moles of CaCO3 produce 2 moles of Ca3SiO5.

Therefore, the mass of limestone required to make 25 kg of cement can be calculated as:

  • Mass of CaCO3 = (Number of moles of Ca3SiO5 x 4) x (Molar mass of CaCO3)
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