4. (2 points) how many grams of oxalic acid dihydrate h2c2o4 -> 2h2o, a diprotic acid, are needed to react with 20.00 ml of 0.4500 m naoh.

The  number of grams  of  oxalic  acid  dihydrate  that are  needed  to react  with  20.00  ml  of 0.4500 M  NaOH is   0.567 grams

    calculation

 Step 1:   write the equation for reaction

H2C2 O4 . 2H2O + 2 NaOH →  Na2C2O4  + 4 H2O

Step 2:   calculate  the  moles of NaOH

 moles =  molarity  x  volume in liters

    volume in liters =  20.00/1000 =0.0200 L

   molarity  =0.4500 M =  0.4500  mol/l

moles = 0.4500 mol/l  x  0.0200 L  =0.0090  moles

Step  3: use the  mole ratio  to  calculate the  of oxalic acid  dihydrate

that is H2 C2O4: NaOH  is 1:2  therefore the moles  of H2C2O4.2H2O

   =0.0090 moles x 1/2 =0.0045   moles

Step  4:  calculate  the  mass  of  H2C2O4.2H2O

 mass= moles x  molar  mass

the  molar mass of H2C2O4.2H2O = (1x2)+ (12 x2) + (16 x4)  + 2(18)  =126 g/mol

mass  =0.0045  moles  x 126  g/mol =0.567 grams

did anyone actually do this? like if anyone has submitted this i rly need .