Stoichiometry & Molar Concentrations

KEY Stoichiometry & Molar Concentrations Name - ______________ 1.) A student wants to put 50.0 ‫ ܮ‬of hydrogen gas at STP into a plastic bag by reacting excess aluminum metal with 3.00 ‫ ܯ‬of sodium hydroxide solution according to the reaction below. What volume of NaOH solution is required? _2_ Al Answer - (s) + _2_ NaOH 50.0 ‫ܪ ܮ‬ଶ × (aq) ଵ ௠௢௟ ுమ ଶଶ.ସଵ ௅ ுమ × + _2_ H2O (l) ଶ ௠௢௟ ே௔ைு ଷ ௠௢௟ ுమ × _2_ NaAlO2  ଵ ௅ ே௔ைு ଷ.଴଴ ௠௢௟ ே௔ைு (aq) + _3_ H2 (g) = 0.496 ‫ܪܱܽܰ ܮ‬ 2.) What volume of 0.250 ‫ ܯ‬HCl is required to completely neutralize 25.0 ݉‫ ܮ‬of 0.318 ‫ ܯ‬NaOH? (Hint balanced equation?) Answer - HCl + NaOH 0.025 ‫× ܪܱܽܰ ܮ‬ ଴.ଷଵ଼ ௠௢௟ ே௔ைு ଵ.଴଴ ௅ ே௔ைு NaCl + H2O  × ଵ ௠௢௟ ு஼௟ ଵ ௠௢௟ ே௔ைு × ଵ ௅ ு஼௟ ଴.ଶହ଴ ௠௢௟ ு஼௟ = 0.0318 ‫݈ܥܪ ܮ‬ 3.) A technician analyzes a sample of water from the “tailings” pond of a mine for the presence of mercury. After treating and concentrating the water sample, the technician carries out the titration reaction found below. A 25.0 ݉‫ ܮ‬sample of the water containing mercury reacts with 15.4 ݉‫ ܮ‬of 0.0148 ‫ ܯ‬Cl-. Hg+2 (aq) + 2 Cl- HgCl2  (aq) (s) a.) What is the molar concentration of the mercury in the water sample? Answer - 0.0154 ‫× ି ݈ܥ ܮ‬ ଴.଴ଵସ଼ ௠௢௟ ஼௟ష ଵ.଴଴ ௅ ஼௟ ష × ଵ ௠௢௟ ு௚శమ ଶ ௠௢௟ ஼௟ ష × ଵ ଴.଴ଶହ ௅ = 0.00456 ‫݃ܪ ܯ‬ାଶ b.) What mass of HgCl2 is formed in the reaction? Answer - 0.025 ‫× ܮ‬ ଴.଴଴ସହ଺ ௠௢௟ ு௚శమ ଵ.଴଴ ௅ × ଵ ௠௢௟ ு௚஼௟మ ଵ ௠௢௟ ு௚శమ × ଶ଻ଵ.ସଽ଺ ௚ ு௚஼௟మ ଵ.଴଴ ௠௢௟ ு௚஼௟మ = 0.0310 ݃ ‫݈ܥ݃ܪ‬ଶ 1 KEY 4.) A 0.10 ݉‫ ܮ‬sample of saturated solution of Ca(OH)2 is reacted with 23.5 ݉‫ ܮ‬of 0.0156 ‫ ܯ‬HCl. a.) What is the molarity of the Ca(OH)2 in the saturated solution? Answer - 2 HCl (aq) + Ca(OH)2 (aq) 0.0235 ‫× ݈ܥܪ ܮ‬ CaCl2 (aq) + 2 H2O (l)  ଴.଴ଵହ଺ ௠௢௟ ு஼௟ × ଵ௅ ଵ ௠௢௟ ஼௔ሺைுሻమ ଶ ௠௢௟ ு஼௟ × ଵ = 1.833 ‫ܯ‬ ଴.଴଴଴ଵ଴ ௅ 1.8 ‫ܽܥ‬ሺܱ‫ܪ‬ሻଶ b.) What mass of Ca(OH)2 is dissolved in 250.0 ݉‫ ܮ‬of saturated Ca(OH)2 solution? Answer - 0.2500 ‫ܽܥ ܮ‬ሺܱ‫ܪ‬ሻଶ × ଵ.଼ଷଷ ௠௢௟ ஼௔ሺைுሻమ ଵ௅ × ଻ସ.ଵ଴ ௚ ஼௔ሺைுሻమ ଵ ௠௢௟ ஼௔ሺைுሻమ = 33.956 ‫ܯ‬ 5.) A student titrates a 2.00 ݉‫ ܮ‬sample of hydrogen peroxide solution, H2O2 2 MnO4- (aq) + 5 H2O2 (aq) + 6 H+ 2 Mn+2  (aq) + 5 O2 (aq) (aq), (g) 34 ݃ ‫ܽܥ‬ሺܱ‫ܪ‬ሻଶ according to the reaction + 8 H2O (l) The supply bottle of hydrogen peroxide is labelled as “3.00% by volume” (3.00 ݉‫ ܮ‬of H2O2 per 100 ݉‫ܮ‬ of solution), which the student calculates to have [H2O2] = 1.24 ‫ܯ‬. a.) What volume of 0.0496 ‫ ܯ‬MnO4- is required for the titration? Answer - 0.002 ‫ܪ ܮ‬ଶ ܱଶ × b.) What volume of O2 Answer - (g) ଵ.ଶସ ௠௢௟ ுమ ைమ ଵ௅ × ଶ ௠௢௟ ெ௡ைర ష ହ ௠௢௟ ுమ ைమ × ଵ௅ ଴.଴ସଽ଺ ௠௢௟ ெ௡ைర ష = 0.0200 ‫ܱ݊ܯ ܮ‬ସ ି at STP is produced during the reaction? 0.0200 ‫ܱ݊ܯ ܮ‬ସ ି × ଴.଴ସଽ଺ ௠௢௟ ெ௡ைర ష ଵ௅ × ହ ௠௢௟ ைమ ଶ ௠௢௟ ெ௡ைర ష × ଶଶ.ସଵ ௅ ைమ ଵ ௠௢௟ ைమ = 0.0556 ‫ܱ ܮ‬ଶ 6.) A 1.00 ݉‫ ܮ‬sample of pure phosphoric acid, H3PO4, is titrated with 43.8 ݉‫ ܮ‬of 0.853 ‫ ܯ‬NaOH according to the reaction _2_ NaOH + ___ H3PO4 ___ Na2HPO4 + _2_ H2O  a.) What is the molar concentration of pure H3PO4? Answer - 0.0438 ‫× ܪܱܽܰ ܮ‬ ଴.଼ହଷ ௠௢௟ ே௔ைு ଵ ௅ ே௔ைு × ଵ ௠௢௟ ுయ ௉ைర ଶ ௠௢௟ ே௔ைு × ଵ ଴.଴଴ଵ଴଴ ௅ = 18.68 ‫ܪ ܯ‬ଷ ܱܲସ 18.7 ‫ܪ ܯ‬ଷ ܱܲସ b.) Calculate the density of pure H3PO4. Answer - ଵ଼.଺଼ ௠௢௟ ுయ ௉ைర ଵ ௅ ுయ ௉ைర × ଽ଼.଴଴ ௚ ுయ ௉ைర ଵ ௠௢௟ ுయ ௉ைర = ଵ଼ଷ଴.଺ସ ௚ ଵ௅ ‫ܪ‬ଷ ܱܲସ ଵ଼ଷଵ ௚ ଵ௅ ‫ܪ‬ଷ ܱܲସ 2 KEY +2 7.) The iron present in a sample of iron ore is converted to Fe Cr2O7-2 + 6 Fe+2 + 14 H+ and titrated with the dichromate ion 2 Cr+3 + 6 Fe+3 + 7 H2O  When 17.6 ݉‫ ܮ‬of 0.125 ‫ ܯ‬dichromate ion is required to react a 25.0 ݉‫ ܮ‬sample of Fe+2 solution, a.) What is the molarity of the Fe+2? Answer - 0.0176 ‫ݎܥ ܮ‬ଶ ܱ଻ ଶି × ଴.ଵଶହ ௠௢௟ ஼௥మ ைళ మష ଵ ௅ ஼௥మ ைళ మష × ଺ ௠௢௟ ி௘ మశ ଵ ௠௢௟ ஼௥మ ைళ మష × ଵ ଴.଴ଶହ଴ ௅ = 0.528 ‫ ݁ܨ ܯ‬ଶା b.) What mass of iron is present in the 25.0 ݉‫ ܮ‬sample? Answer - 0.0250 ‫ ݁ܨ ܮ‬ଶା × ଴.ହଶ଼ ௠௢௟ ி௘ మశ ଵ௅ × ହହ.଼ସହ ௚ ி௘ మశ ଵ ௠௢௟ ி௘ మశ = 0.737 ݃ ‫ ݁ܨ‬ଶା 8.) Prior to analyzing a fertilizer sample containing NH4NO3, a chemist makes a test solution by dissolving 15.5 ݃ of pure NH4NO3 and diluting it to 500.0 ݉‫ܮ‬. If the chemist wishes to carry out the titration reaction ___ NH4NO3 (aq) + ___ NaOH ___ NH3  (aq) + ___ H2O (l) + ___ NaNO3 (g) (aq) such that the reaction requires 25.0 ݉‫ ܮ‬of NaOH when 10.0 ݉‫ ܮ‬of the NH4NO3 solution is titrated, a.) What is the molarity of the NaOH they should use? Answer - 0.010 ‫ܪܰ ܮ‬ସ ܱܰଷ × b.) What volume of NH3 Answer - (g) ଵହ.ହ ௚ ேுర ேைయ ଴.ହ଴଴ ௅ ேுర ேைయ × ଵ ௠௢௟ ேுర ேைయ × ଼଴.଴ସଶ଺ ௚ ேுర ேைయ ଵ ௠௢௟ ே௔ைு ଵ ௠௢௟ ேுర ேைయ × ଵ ଴.଴ଶହ ௅ ே௔ைு = 0.155 ‫ܪܱܽܰ ܯ‬ at STP would be produced? 0.010 ‫ܪܰ ܮ‬ସ ܱܰଷ × ଵହ.ହ ௚ ேுర ேைయ ଴.ହ଴଴ ௅ ேுర ேைయ × ଵ ௠௢௟ ேுర ேைయ ଼଴.଴ସଶ଺ ௚ ேுర ேைయ × ଵ ௠௢௟ ேுయ ଵ ௠௢௟ ேுర ேைయ × ଶଶ.ସଵ ௅ ேுయ ଵ ௠௢௟ ேுయ = 0.0868 ‫ܪܰ ܮ‬ଷ 9.) The CO2 content of a 10.0 ‫ ܮ‬sample of air at STP is determined as follows. The air pumped through a flask containing 25.0 ݉‫ ܮ‬of 0.0538 ‫ ܯ‬Ba(OH)2, precipitating the CO2 present as BaCO3 ___ Ba(OH)2 (aq) + ___ CO2 (g)  ___ BaCO3 (s) + ___ H2O (l) a.) How many moles of Ba(OH)2 are present in the original Ba(OH)2 solution? Answer - 0.0250 ‫ܽܤ ܮ‬ሺܱ‫ܪ‬ሻଶ × ଴.଴ହଷ଼ ௠௢௟ ஻௔ሺைுሻమ ଵ௅ = 0.001345 ݃ 0.00135 ݃ ‫ܽܤ‬ሺܱ‫ܪ‬ሻଶ 3 KEY b.) Only a small amount of the Ba(OH)2 present reacts with the added CO2. The remaining unreacted Ba(OH)2 is titrated with hydrochloric acid according to the equation Ba(OH)2 + 2 HCl  BaCl2 + 2 H2O If the titration requires 23.0 ݉‫ ܮ‬of 0.104 ‫ ܯ‬HCl, how many moles of Ba(OH)2 solution after reacting with the CO2 in the air? Answer - 0.0230 ‫× ݈ܥܪ ܮ‬ ଴.ଵ଴ସ ௠௢௟ ு஼௟ ଵ௅ × ଵ ௠௢௟ ஻௔ሺைுሻమ ଶ ௠௢௟ ு஼௟ = 0.001196 ݉‫ܽܤ ݈݋‬ሺܱ‫ܪ‬ሻଶ 0.00120 ݉‫ܽܤ ݈݋‬ሺܱ‫ܪ‬ሻଶ c.) How many moles of Ba(OH)2 are reacted by the CO2? Answer - 0.001345 − 0.001196 = 0.000149 ݉‫ܽܤ ݈݋‬ሺܱ‫ܪ‬ሻଶ d.) How many moles of CO2 are in the sample of air? Answer - 1 to 1 ratio so . . . . . 0.000149 ݉‫ܱܥ ݈݋‬ଶ e.) How many litres of CO2 at STP are contained in the 10.0 ‫ ܮ‬sample of air? What percentage of the air sample’s volume is CO2? Answer - 0.000154 ݉‫ܱܥ ݈݋‬ଶ × Answers – 1.) 2 Al (s) + 2 NaOH (aq) + 2 H2O (l) 2.) HCl + NaOH  NaCl + H2O ଶଶ.ସ ௅ ଵ ௠௢௟ ஼ைమ  0.0318 ‫ܮ‬ = 0.00345 ‫ܱܥ ܮ‬ଶ 2 NaAlO2 (aq) + 3 H2 3a.) 0.00456 ‫ܯ‬ 4.) Ca(OH)2 + 2 HCl  CaCl2 + 2 H2O a.) 1.83 ‫ ܯ‬b.) 33.9 ݃ 6.) 2 NaOH + H3PO4  Na2HPO4 + 2H2O a.) 18.7 ‫ ܯ‬b.) 1831 9a.) 0.00135 ݉‫ ݈݋‬b.) 0.00120 ݉‫ ݈݋‬c.) 0.000149 ݉‫݈݋‬ d.) 0.000149 ݉‫݈݋‬ (g) 0.496 ‫ܮ‬ b.) 0.0310 ݃ 5a.) 0.0200 ‫ ܮ‬b.) 0.0556 ‫ܮ‬ ௚ ௅ 7a.) 0.528 ‫ ܯ‬b.) 0.737 ݃ 8a.) 0.155 ‫ ܯ‬b.) 0.0868 ‫ܮ‬ e.) 0.00345 ‫ܮ‬ 4