Asked by Anonymous
Determine how many milliliters of 0.448 M HClO4 will be required to neutralize 38.30 g Ca(OH)2 according to the reaction: 2HCl+Ca(ClO4)2+2H20
Answers
Answered by
Dr Russ
You need to start from a balanced equation
HClO4 + Ca(OH)2 -> Ca(ClO4)2 + H20
which you need to balance
2HClO4 + Ca(OH)2 -> Ca(ClO4)2 + 2H20
then work out how many moles 38.30 g Ca(OH)2 represents from
number of moles = 38.30 g/molar mass
from the equation 2 mole of HClO4 needed for each mole of Ca(OH)2
hence calcutae the number of moles of HClO4 needed.
ml of HClO4 needed is then
number of moles x 1000/(0.448 mole L^-1)
HClO4 + Ca(OH)2 -> Ca(ClO4)2 + H20
which you need to balance
2HClO4 + Ca(OH)2 -> Ca(ClO4)2 + 2H20
then work out how many moles 38.30 g Ca(OH)2 represents from
number of moles = 38.30 g/molar mass
from the equation 2 mole of HClO4 needed for each mole of Ca(OH)2
hence calcutae the number of moles of HClO4 needed.
ml of HClO4 needed is then
number of moles x 1000/(0.448 mole L^-1)
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