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
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).
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