How man moles of H2C2O4*2H2O is required t completely neutralize 25 mL of .15M NaOH?

Answer 1

25mL of 0.15M correspond to 0.15*25*10^-3 =0.00375Moles of NaOH
In the reaction of C2H2O4,2H2O+NaOH-->Na2COO2+4H2O
you need two moles of NaOH for 1 of H2C2O4
so for 0.00375moles of NaOH you need0.00375/2=0.001875mole=1.875*10^-3 moles or 1.875mM

Answer 2

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Answer 3

Ba(OH)2(aq) + 2HCl(aq) ---> BaCl2(aq) + 2H2O(l) First we verify the moles of barium hydroxide that reacted with the HCl by using multiplying the quantity (in liters) by using the concentration n(Ba(OH)2) = (0.0.5 L)(0.15 M) = 3.75x10^-3 mol next verify the moles of HCl employing the mole ratio; 2 moles of HCl reacts with a million mole of Ba(OH)2. So the moles of HCl is two cases the moles of Ba(OH)2: n(HCl) = 2 x n(Ba(OH)2) = (2)(3.75x10^-3) = 7.50x10^-3 mol finally we verify the quantity of HCl required by using employing the equation for molarity: Molarity = moles / quantity quantity = 7.50x10^-3 mol / 0.seventy 5 M = a million.00x10^-2 L = 10.0 mL