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Determine how many milliliters of 0.626 M HClO4 will be required to neutralize 57.90 g Ca(OH)2 according to the reaction
2 HClO4(aq) +Ca(OH)2(S) =Ca(ClO4)2(aq) + 2 H2O(l)
HCLO4 106 g/mol
CaOH2 74.1 g/mol
You need to find out how many moles you need. let's look at
57.9g x (1 mol CaOH2/74.1g) = 0.78 mol of CaOH2 This is how many moles you have
but you need to find out how many moles of something will react completely with this so
you find the # of moles of HCLO4 by multiplying by the molar ratio.
0.78mol CaOH2 x (2mol HClO4/1 CaOH2) = 1.56 mol. HClO4.
so 0.78 mol of CaOH2 is needed to react with 1.56 mol of HCLO4.
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found a useful site that might help out http://www.chemteam.info/AcidBase/Titration-calc-amount-solid2.html
according to what was done there:
to find the original volume in a neutralization problem you need to do:
(x) = volume
mol/L (x) = moles
moles/(mol/L) gives you Liters required.
we have 0.626 mol/L of HClO4 and we needed 1.56 mol of that to complete the reaction.
so it's (1.56mol)/(0.626moL/L) = 2.50 L needed.