Question
When a reaction mixture with a total volume of 1390 mL that is 0.00791 M aqueous Ba2+ was stoichiometrically produced as per the balanced equation, what mass (g) of Ba2+ was required?
Ba(OH)2(aq) + 2 HClO4(aq) → Ba(ClO4)2(aq) + 2 H2O(l)
.00791x1.39=mole of Ba2+
mole of Ba2+ xmolecular weight=grams Ba2+
Ba(OH)2(aq) + 2 HClO4(aq) → Ba(ClO4)2(aq) + 2 H2O(l)
.00791x1.39=mole of Ba2+
mole of Ba2+ xmolecular weight=grams Ba2+
Answers
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