Reaction Stoichiometry and Solution Concentration Q1. FeS(S) + 2HCl(aq) FeCl2(S) + H2S(g) Q2. C6H10(g) + O2(g) CO2(g) + H2O(g) Q3.

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1 Reaction Stoichiometry and Solution Concentration Q1. The reaction between Iron(II) sulfide and HCl is as follows; FeS (S) + 2HCl (aq) FeCl 2(S) + H 2 S (g) What will be the number of moles of each reactant left if initially there were 8.9moles of FeS and 6.3 moles of HCl. Q2. Following is the combustion reaction of hexyne. C 6 H 10(g) + O 2(g) CO 2(g) + H 2 O (g) After combusting 5.63g of hexyne with 15.5g of oxygen 10.24g of CO2 gas obtained. Answer the following; a). Write the balanced chemical equation b). Which one is the limiting reactant? c). The excess reactant? d). What is the theoretical yield for CO2? e). Percentage of yield of the reaction? f). Amount of excess reactant left over. Q3. For the following reaction between Calcium and oxygen, if 85.2g of Ca were consumed how much grams of oxygen were reacted? Ca (s) + O 2(g) CaO (s) Q4. The neutralization reaction between sulfuric acid and sodium hydroxide results in the formation of water and sodium sulfate. a). Write the balanced chemical equation for this aqueous reaction. b). How many moles of sodium hydroxide will be needed for the complete neutralization of 9.46 moles of sulfuric acid Q5. What is the percentage of yield if the reaction between 8.9g of nitrogen gas and 4.6 g of hydrogen gas produced 8.4g of ammonia? Q6. How much grams of lithium nitrate must be dissolved to get a 500.0mL 1.5M aqueous solution? Q7. What will be the concentration of the aqueous solution prepared by dissolving 6.50g of calcium carbonate in water and made it up to 250mL. Q8. What will be the final concentration if 45mL of 2.5M HCl is diluted up to 150mL?

2 Q9. The number of moles of chloride ions present in 250mL of 1.2mM aqueous calcium chloride solution will be? Q10. What is the molarity of ClO3 - in the following aqueous solutions; a). 1.25M Ca(ClO3)2 b). 0.56M Al(ClO3)3 c). 0.02M HClO3 Q11. How much volume of 0.2M Ca(OH)2 (aq)will be needed for the complete neutralization of 20.0mL of 0.5M, H2SO4(aq) Ca(OH) 2 (aq) + H 2 SO 4(aq) CaSO 4(S) + 2H 2 O (l) Q12. The volume of 0.50M sodium hydroxide (NaOH) stock solution you must take to dilute it to 500mL, 0.15M NaOH solution? Q mL of 0.25M PbNO3 reacted with 75.0mL of 0.1M HCl. The reaction is as follows; Pb(NO 3) 2(aq) + 2HCl (aq) PbCl 2(S) + 2H NO 3 (aq) If the mass of PbCl2 formed were 0.86g. What is the percentage of yield of the reaction? Reference ; Tro, Nivaldo J. Chemistry: A Molecular Approach., Print.

3 Answers 1. 0 moles HCl, 5.75 moles FeS Moles of FeS needed to react with 6.3 moles of HCl = 6.3 moles of HCl 1 moles of FeS 2 moles of HCl = 3.2 moles of FeS Limiting Reactant =HCl (completely use up) FeS left = = 5.7 moles 2. a). 2 C 6 H 10(g) + 17 O 2(g) 12 CO 2(g) + 10 H 2 O (g) b). Amount of CO2 can be maded from given reactant; 5.63 g C 6 H 10 1 moles of C 6H gc 6 H moles of CO 2 2 moles of C 6 H g CO 2 1 moles CO 2 = 18.1 g CO g O 2 1 moles of O go 2 c). Hexyne d) g 12 moles of CO 2 17 moles of O 2 Limiting Recatant = Oxygen g CO 2 1 moles CO 2 = g CO 2 Theoretical Yield e). % of Yield = Actual Yield Theoretical Yield 100 = = 68.0 % f). Amount of Hexyne reacted g O 2 1 moles of O go 2 Amount of hexyne left = 0.95 g 3. Balanced chemical equation; 2Ca (s) + O 2(g) 2CaO (s) 2 moles of C 6H moles of O gc 6H 10 1 moles C 6 H 10 = 4.68 g C 6H g Ca 1 moles of Ca g Ca 1 moles of O 2 2 moles of Ca go 2 1 mole O 2 = g O g of O 2 4. a). 2NaOH (aq) + H 2 SO 4(aq) Na 2 SO4 (aq) + 2H 2 O (l)

4 b) moles 9.46 moles H 2 SO 4 5). N 2 (g) + 3H 2(g) 2NH 3 (g) 2 moles of NaOH 1 moles of H 2 SO 4 = 18.9 moles of NaOH 8.9 g N 2 1 moles of N gn 2 2 moles of NH 3 1 moles of N g NH 3 1 moles NH 3 = 21.6 g NH 3 - Theoretical Yield 4.6 g H 2 1 moles of H g H 2 % Yield = 8.4/21.6 *100 = 39% 6). LiNO 3 (68.95 g/mol) 500 mllino 3 2 moles of NH 3 3 moles of H g NH 3 1 moles NH 3 = 25.8 g NH ml 1.5 moles LiNO g LiNO 3 = 52 glino 1 moles LiNO 3 3 7) g CaCO 3 1.molesCaCO g CaCO ml 1000 ml 1L = 0.26 M 8). M 1 V 1 = M 2 V M 9). 250 ml CaCl ml moles CaCl 2 2 moles Cl 1 moles CaCl 2 = ). a).2.5 M b) M c). 0.02M 11) ml 0.02 L H 2 SO moles of H 2SO 4 1 moles of Ca(OH) 2 1 moles of H 2 SO moles Ca(OH) 2 = 0.05 L Ca(OH) 2 = 50.0 ml 12). M 1 V 1 = M 2 V ml 13). Pb(NO 3) 2 = g/mol, HCl = g/mol and PbCl 2 = g/mol

5 0.25 Moles of Pb(NO 3)2 1 moles of PbCl L Pb(NO 3)2 1 moles of Pb(NO 3) g PbCl2 = 3.48 g PbCl2 1 moles PbCl L HCl 0.1 Moles of HCl 1 moles of PbCl2 2 moles of HCl g PbCl2 = 1.04 g PbCl2 1 moles PbCl2 % 0f yield = (0.86/1.04)*100=82.7%