Download Chemical Equilibrium and Reaction Calculations and more Assignments Chemistry in PDF only on Docsity! 1. my numbers are 1.21g of compound burned in excess 02, forming 1.05g water XaHb + 02 H20 + ? The molar ratio of XaHb : H20 = 2/b 1.21g compound/ 62.09 g/mol= 0.0195 mol compound 1.05g H20/18 g/mol= .05833 mol water 2/b= 0.0195/.05833; b=6 XaHb has a mass of 62.09, so a moles of X = 56g X=14 (N) a=4, fits best so the formula is N4H6 2. find mass of 2% NH4 ions by mass 10000 kg waste x (3 kg NH4/ 100 kg waste) x (1000g/kg) x (1 mol NH4/ 18 g NH4) x (1 mol C5H702N/ 55 mol NH4) x (113.1 g C5H702N/ mol C5H702N) = 23000g if 2& NH4 by mass 22000g if 98% converted 3. a) .350 mol Ca(NO3)2 in 100mL soln Ca2+= 3.50 M NO3- = 7 M b) 6.5 mol Na2SO4 in 1.25 L of solution Na+ = 1M SO4 2-= 5.2 M c) 5.00 g of NH4Cl in 800.0 mL of solution NH4+ = .117M Cl-= .117 M d) 1.00 g K3PO4 in 400.0 mL of solution K+=.0353 PO43-= .0118 4. 18 mg dissolved stock soln= 18mg/ 500mL= .018 g/ 500 mL= 3.6 x 10E -5 g steroid 100x 10E-6 L stock soln x (1000 mL/L) x 3.6x10E-5 g steroid/mL)= 3.6 x10E-6 g steroid dilution: 3.6x10E-6 g steroid/100mL x (1000mL/L)x (1 mol steroid/ 334 g)= 1.07 x10E-7M 5. What volume of 0.850 M Na3PO4 is required to precipitate all the lead(II) ions from 120.0 mL of 0.550 M Pb(NO3)2? 2 Na3PO4 + 3 Pb(NO3)2 Pb (PO4)2 + 6 Na(NO3) .120 L x (0.550 mol Pb(NO3)2 /L) x 2 mol Na3PO4/3 mol Pb(NO3)2) x 1L Na3PO4/ .085 mol Na3PO4)= 52 mL 6. HA +NaOH NaA + HsO mol HA= .0250 L x (.500mol NaOH/ L) x (1 mol HA/mol NaOH)= 0.0125 mol HA x g HA/mol HA= 2.20g HA/0.0125 mol x= molar mass HA= 176 g empirical formula C3H4O3= 88 g/mol 176g/mol / 88 g/mol= 2 so molecular formula is C6H8O6 7. a)-1 b) +1 c)0 d) +7 e) +3 8. (a) 2 C2H6(g) + 7 O2(g) 4CO2(g) + 6 H2O(g) (b) 1 Mg(s) + 2 HCl(aq) 1 Mg2+(aq) + 2 Cl -(aq) + 1 H2(g) (c) 1 Cu(s) + 2 Ag+(aq) 1 Cu2+(aq) + 2 Ag(s) (d) 1 Zn(s) + 1 H2SO4(aq) 1 ZnSO4(aq) + 1 H2(g) 9. 1Au(s) + 4Cl -(aq) + 4H+(aq) + 1[[NO3]-](aq) 1 [[AuCl4]-](aq) + 1 [NO](g) + 2 [H2O](l) 10. How many grams of silver chloride can be prepared by the reaction of 121.9 mL of 0.22 M silver nitrate with 121.9 mL of 0.14 M calcium chloride? 2 AgNO3 + CaCl2 2 AgCl + Ca(NO3)2 mol AgNO3= .1000 L x (. 22 mol AgNO3/ L) = 0.022 mol mol CaCl2 = .1000L x (0.14 mol CaCl2/L) = 0.014 mol they react in a ratio of 2:1 from the balanced eqn, but the real mole ratios are 0.022/.014= 1.6, so AgNO3 is limiting reagent
