Question

5.(25pts) The Ca' in a sample of limestone weighing 0.142g was precipitated by using oxalate ions C204 to produce Ca C2O4(s) via the reaction Ca²+ (aq) + C3042 (aq) -----> Ca C2O4(s) . If the resulting Ca C2O4(s) weighed 0.153 g, what was the % Ca'' in the original sample? Bonus Problem (+6pts. No partial credit. A 500.0 mg table of an antacid containing Mg(OH)2, Al(OH)3, and a binder was dissolved in 50.0 mL of 0.500 M HCL. The resulting solution was acidic, and it needed 30.9 mL of 0.255 M NaOH for neutralization. (a) Calculation the number of moles of OH' ions in the tablet. (b) If the tablet contains 5.0% binder, how many milligrams of Mg(OH)2 and how many of Al(OH)3 does the tablet contain?

Answer 1

mass Massi 0.1539 5 Ca²+ + Gout (a Coy Resulting solution acidie Molar: 40 glool 88 g/mol 128 g/mol it required 30.9mL of 0.255m. NaoH According to log 88g 128g. 30.9 mK x 0.255 may Stoichiomeboy (1x40) (x86) (lxp B) - 7.8795 mmol of Naok HCl + NaOH - Naci+H2O Asit is clear from equation i ri 1 mol of Nach neutralise 1 mol of HOI Sy is produ 9409 (62+ 1 7.8795 mmol of Naot will neutralise 7.8795 mmol of HCl : 0.153g . " 40 x 0.1539 128 left HCl = 25-7-8795 mass of c2t = 0.0478125 g = 1.7. 1205 mmol of HCl in Sample . l atin Sample = mass of Ca²+ xloo & Left HCl which was initially total sample mass present and made to react with antacid tablet contained = 0.0478125 g xloo 17:1205 mmol of HCl 0.142 g Acid Reacted with Antacid = 133.67./ / 17:1205 monol of HCl : 33.67% of Ca²+ was present is the Thus Antacid must be containing Sample 17.1205 mmol of OH - Molo & a) Thus, 17.1205.00 5ooong tablet - Alloh Bender I 10.0171205 mol of Oht Res tablet was dissolved in som of 0.5MHCl were present in oft tablet 50 mL x 0.5 ml - 25 mmol HCl (6) Since; Binder is 5%. & [ A(OH)3] + Mg(OH)2 = 95%

= 951 of 500mg = 95 5* = 47sing mass (mycom) +Aicon)) = 475 mg n AOH)₂ AB+ + 3n OH m Mg(OH) → m Max + 2moh het n= no of moles of Allohla m = no. of moles of Malohl, Also, we have (OH) total = 0.0171205 Mol Molar mass Al(OH)₂ = 78 g/mol "Mg(OH)2 = 58.32 g/mol Erol This two equations can be male; mass = molar mass x moles n 78000 tm 58320 = 475 mg 2723 and 3n + 2m = 0.017/205 mol anol -006848 modlar mass (g/mol) has been conerted into mg/mol Multiplying ♡ with 3 and ② with 78000 we get MOL 234006 n +174960 m =1425 234 boon t 156000 m = 1335.4 18960 m = 89.6

m= 4.72574x10 moles @ Putting values of m into and solving for ni n = 2.5563 810 3 moles Thus, moles of Allot z = 2.5563x16 modes mass "= 2.5563 x 103 x 7 8 9 mol mol = 0.19939 g mass AlloH)₃ = 199.ung ( 1000 mg = 19) moles of Mg(OH)₂ = 4.725 74 X163 mass " " = 4.72574 x 3 x 58.32g mol MOL = 0.2756059 mass MaloH)₂ = 27506 mg Thus amount of Allowly and malontz in the tablet were calculated.