1 / 7

Answers to Volumetric Calculations II

Answers to Volumetric Calculations II. Question 1 2NaOH + H 2 SO 4  Na 2 SO 4 + 2H 2 O No. of mol of NaOH = 20.0/1000 x 0.200 = 0.00400 2 mol of NaOH react with 1 mol of H 2 SO 4 (Mole ratio) No. of mol of H 2 SO 4 reacted = 0.00400/2 = 0.00200

kelsie-kidd
Download Presentation

Answers to Volumetric Calculations II

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Answers to Volumetric Calculations II Question 1 2NaOH + H2SO4 Na2SO4 + 2H2O No. of mol of NaOH = 20.0/1000 x 0.200 = 0.00400 2 mol of NaOH react with 1 mol of H2SO4 (Mole ratio) No. of mol of H2SO4 reacted = 0.00400/2 = 0.00200 Sulfuric acid is a dibasic acid. H2SO4  2H+ + SO42- 1mol of H2SO4 gives 2 mol of H+ No. of mol of H+ = 0.00200 x 2 = 0.00400 Hence concentration of H+ = 0.00400 ÷ (32.0/1000) = 0.125 mol dm-3( 3 sf !!!)

  2. Question 2 CaCO3 + 2HCl  CaCl2 + H2O + CO2 No. of mol of CaCO3 = 5/(40+12+3x16) = 0.0500 1 mol CaCO3 reacts with 2 mol HCl 0.0500 mol CaCO3 react with 0.100 mol HCl No. of mol = volume in dm3 x conc in mol/dm3 0.100 = Volume x 0.400 Volume = 0.250 dm3 or 250 cm3 ( 3 sf !!!)

  3. Question 3 No. of mol of H3PO4 in 26.25 cm3 = 26.25/1000 x 0.0400 = 1.05 x 10-3 No. of mol OH- in 1 dm3 = 1.80/(16+1) = 0.1058 NaOH  Na+ + OH- 1 mol of OH- is produced by 1 mol of NaOH Hence no. of mol of NaOH in 1 dm3 = 0.1058 No. of mol of NaOH in 20.0 cm3 =20/1000 x 0.1058 = 2.116 x 10-3 1.05 x 10-3 mol H3PO4 react with 2.116 x 10-3 mol NaOH. Hence 1 mol H3PO4 reacts with 2.116 x 10-3 / 1.05 x 10-3 = 2.02 mol of NaOH

  4. Question 3 • From part (a), 1 mol H3PO4 : 2.02  2 mol NaOH • Use this information to write equation • H3PO4 + 2NaOH  Na2HPO4 + 2H2O

  5. Question 4 • Precipitation (NOT displacement) • CuSO4 + 2NaOH  Cu(OH)2 + Na2SO4 • No. of mol of CuSO4 = 10.0/1000 x 1.00 = 0.0100 • No. of mol of NaOH = 10.0/1000 x 0.100 = 0.00100 • 2 mol of NaOH react with 1 mol of CuSO4 • 0.00100 mol of NaOH react with 0.00100 ÷2 • = 0.000500 mol of CuSO4 • But the initial amt of CuSO4 is 0.0100 mol, hence CuSO4 is in excess and NaOH is the limiting reactant.

  6. Question 4 (Cont’d) b) Use the limiting reactant to calculate mass of Cu(OH)2 2 mol of NaOH produce 1 mol of Cu(OH)2 0.00100 mol of NaOH produce 0.000500 mol of Cu(OH)2 Mass of Cu(OH)2 = 0.000500 x [64+2(16+1)] = 0.0490 g ( 3 sf !!!)

  7. *Question 5 HCl + NaOH  NaCl + H2O No. of mol NaOH = 35.0/1000 x 0.200 = 7.00 x 10-3 1 mol NaOH reacts with 1 mol HCl Hence no. of mol of HCl remaining= 7.00 x 10-3 No. of mol HCl initially = 50.0/1000 x 0.500 = 0.0250 No. of mol HCl reacted = 0.0250 - 7.00 x 10-3 = 0.0180 CaCO3 + 2HCl  CaCl2 + H2O + CO2 2 mol HCl react with 1 mol CaCO3 Hence 0.0180 mol HCl react with 9.00 x 10-3 mol CaCO3 Mass of CaCO3 = 9.00 x 10-3 x (40+12+3x16) = 0.900 g % purity = 0.900/1.00 x 100 = 90.0 % ( 3 sf !!!)

More Related