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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?

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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

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