calculate how many moles of CaCl2•2H2O are present in 1.50 g of CaCl2•2H2O and then calculate how many moles of pure CaCl2 are present in the 1.50 g of CaCl2•2H2O.

This is the calcium chloride dihydrate (Because there are 2 molecules of water). That means you have 1 atom of Ca, 2 atoms of Cl, 4 atoms of H and 2 atoms of C. Molar mass is the algebraic sum of the atomic weights that compose the molecule.

Molar mass:

Ca: 1ₓ40.08g/mol = 40.08g/mol

Cl: 2ₓ35.45g/mol = 70.90g/mol

H: 4ₓ1.01g/mol: 4.04g/mol

O: 2ₓ16g/mol: 32g/mol.

40.08 + 70.90 + 4.04 + 32 = 147.02g/mol

By using the molar mass of the susbtance, 1.50g are:

1.50g ₓ (1mol / 147.02g) =

0.0102moles CaCl₂.2H₂O

In the 1.50g of CaCl₂.2H₂O you have 0.0102 moles. As you can see, 1 mole of the dihydrate substance contains 1 mole of CaCl₂. Thus, you have:

0.0102moles CaCl₂

nikitarahangdaleVT nikitarahangdaleVT · Answer 2 · 2022-04-06T12:32:18+02:00

The moles of CaCl₂•2H₂O and pure CaCl₂ present in the 1.50 grams of CaCl₂•2H₂O is 0.0102 moles.

How we calculate moles?

Moles of any substance will be calculated as:

n = W/M, where

W = given mass

M = molar mass

In the question it is given that,

Mass of CaCl₂•2H₂O = 1.50 g

Molar mass of CaCl₂•2H₂O = 147.02 g/mole

Moles of CaCl₂•2H₂O = 1.50 / 147.02 = 0.0102 moles

On heating the given compound we get,

CaCl₂•2H₂O → CaCl₂ + 2H₂O

From the stoichiometry of the reaction it is clear that in one moles of the dihydrate compound, one mole of CaCl₂ is present.

So, in 0.0102 moles of CaCl₂•2H₂O, same moles of CaCl₂ will present.

Hence, in 1.50g of CaCl₂•2H₂O, 0.0102 moles of CaCl₂•2H₂O and pure CaCl₂ is present.

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https://brainly.com/question/24631381