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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)
57.90 g = 1 mol / 74.904 = 2 mol / 1 mol, right? :/
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.

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