Question

A 66.4 gram sample of Ba(ClO4)2 3 H2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained?

Answer 1

Answer:

57.2 g

Explanation:

First we convert 66.4 grams of Ba(ClO₄)₂·3H₂O into moles, using its molar mass:

• Molar mass of Ba(ClO₄)₂·3H₂O = Molar mass of Ba(ClO₄)₂ + (Molar Mass of H₂O)*3

• Molar mass of Ba(ClO₄)₂·3H₂O = 390.23 g/mol

• 66.4 g ÷ 390.23 g/mol = 0.170 mol Ba(ClO₄)₂·3H₂O

0.170 moles of Ba(ClO₄)₂·3H₂O would produce 0.170 moles of 0.170 moles of Ba(ClO₄)₂. Meaning we now convert 0.170 moles of Ba(ClO₄)₂ into grams, using the molar mass of Ba(ClO₄)₂:

• 0.170 mol * 336.23 g/mol = 57.2 g