How many grams of Ca2+ in 42.51 g of limestone (CaCO3)? Answer to 2 decimal pointsMW (Ca) = 40.08 g.mol-1MW (CaCO3) = 100.09 g.mol-1

Step 1: Identify the given values

The mass (m) of CaCO3 = 42.51 g The molar mass (M) of CaCO3 = 100.09 g/mol The molar mass (M) of Ca = 40.08 g/mol

Step 2: Calculate the number of moles (n) of CaCO3

The formula to calculate the number of moles is:

n = m/M

Substitute the given values into the formula:

n = 42.51 g / 100.09 g/mol = 0.4246 mol

Step 3: Calculate the mass of Ca2+ in the limestone

In CaCO3, there is one mole of Ca for every mole of CaCO3. So, the number of moles of Ca is also 0.4246 mol.

The formula to calculate the mass from the number of moles is:

m = n * M

Substitute the given values into the formula:

m = 0.4246 mol * 40.08 g/mol = 17.01 g

So, there are 17.01 g of Ca2+ in 42.51 g of limestone.