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How many grams of Ca2+ in 81.59 g of limestone (CaCO3)? Answer to 2 decimal pointsMW (Ca) = 40.08 g.mol-1MW (CaCO3) = 100.09 g.mol-1

Question

How many grams of Ca2+ in 81.59 g of limestone (CaCO3)?

Answer to 2 decimal points

  • MW (Ca) = 40.08 g/mol
  • MW (CaCO3) = 100.09 g/mol
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Solution

The number of grams of a component in a compound can be calculated using the formula:

Grams of component = (mass of compound / molar mass of compound) * molar mass of component

Given: Mass of CaCO3 (limestone) = 81.59 g Molar mass of CaCO3 = 100.09 g/mol Molar mass of Ca = 40.08 g/mol

Step 1: Substitute the given values into the formula:

Grams of Ca = (81.59 g / 100.09 g/mol) * 40.08 g/mol

Step 2: Perform the calculation to find the grams of Ca:

Grams of Ca = 32.67 g

So, there are 32.67 grams of Ca2+ in 81.59 g of limestone (CaCO3).

This problem has been solved

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