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Download CHEM 1201 (Basic Chemistry) – Exam 2 Form 3 | LSU Mastery Guide with Expert Solutions and more Exams Chemistry in PDF only on Docsity!
Fall 2024 Exam 2 Form 3 MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) How many molecules of HCI are formed when 60.0 g of water reacts according to the following balanced reaction? Assume excess ICI3. 2 ICI3 + 3 H2O > ICI + HIO3 + 5 HCI 1) A) 3.34 x 1025 molecules HCI B) 4.00 x 1024 molecules HCI C) 6.00 x 1024 molecules HCI D) 2.00 x 1024 molecules HCI E) 3.34 x 1024 molecules HCI 1molH,Q 5SmolHCl 6.022 x 1073 molecules 60.09 H20* 75 04 gH,0° 3molH,0" Imol HCl = 3.34 x 1074molecules HCL 2) A solution is prepared by mixing 5.00 mL of 0.100 M HCI and 2.00 mL of 0.200 M NaCl. What is the molarity of chloride ion in this solution? 2) A)0.174 B) 7.78 [cr] = (Volume(L)of NaCl x M of NaCl) + (Volume(L)of HCL x M of HCl) C) 0.0143 Total Volume (L) D) 0.450 -, (-200 mol Nacl . (100 mol HCL 002 L NaCl (Pe ar) + 0.008L Natl (are —) (.002 L+ .005 L) E) 0.129 [cr] = =0.129M 3) How many grams of CaCl2 are formed when 20.00 mL of 0.00237 M Ca(OH)2 reacts with excess Cl2 gas? 2 Ca(OH)2 (aq) + 2 Cle (g) + Ca(OCl)2 (aq) + CaCl2 (s) + 2 H20 (I) 3) A) 0.0380 g B) 0.0105 g C) 0.00263 g D) 0.00526 g E) none of these Mass of CaCl, 1LCa(OH), 0.00237 moles Ca(OH), _1molCaCl, 110.989 CaCl, 20.00 mL Ca(OH)2 * 7960 mE Ca(OH)a* TL Ca(OH), * Ymol Ca(OH), "Amol cacti, ~ 00763 9 Cache 4) Which of the following is NOT a strong electrolyte? 4) A) MgCO3 B) K2S04 C) CaCl2 D) KOH E) NaC2H302 5) Methane (CHa) and oxygen react to form carbon dioxide and water. What mass of water is formed if 1.6 g of methane reacts with 6.4 g of oxygen to produce 4.4 g of carbon dioxide? 5) A)8.0g a ae CH, + 20, > COz + 2H,0 ot one of these Mass of 10 = 3.6gH,0 1molCH,\ 2molH,0 18.02 g H,0 16g CHy+( mo 4.) abbot gm 16.049 CH,) TmolCH, 1molH,0 1mol 0, 2molH,0 18.02 gH,0 6.4g 02+ ( ) 32.0g0,)" 2mol0,° 1molH,o ~ *°9 #20 Neither is a limiting reagent 6) How many grams of NaOH (MW = 40.0) are there in 500.0 mL of a 0.250 M NaOH solution? 6) A) 14.0 = 5.00 Mass of NaOH = Volume (L) x Molarity (M) x Molar Mass 4) C) 0.00219 mo D) 0.125 1LNaOH .250molNaOH 40.0 g NaOH E) 144 = 500.0 mL NaOH « 1000mLNaOQH 1LNaOH ~* 1molNaOH =5.00g 7) Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaQ(s) + H20 (|) > Ca(OH) (s) A 1.50-g sample of CaO is reacted with 1.45 g of H20. How many grams of water remain after the reaction is complete? yy A) 1.04 B) 0.00 Remaining mass of H,0 Cy 0 o028F 1molCaQ 1molH,0 18.02 gH,0 D) 0.966 mol Cai mol Hz 02 g HO _ E) 0.0536 1.509 Ca0* < agg Ca0* TmolCaO* 1molH,0 ~ 1229 #20 1.45 g H,0—.482 gH,0 =.968 g H,0 11) What volume (mL) of a 3.20 M lead nitrate solution must be diluted to 847.6 mL to make a 2.19 M solution of lead nitrate? A) 1240 B) 580 C) 0.00172 D) 0.00827 E) 5940 12) What mass (in kg) does 4.75 moles of cobalt have? A) 0.820 kg B) 0.280 kg C) 0.632 kg D) 0.122 kg E) 0.352 kg m = 44.75 mol Co * 1) M2V2 1M y, — (2:19 M)(847.6 mL) 1” 3.20M V, = 580.mL 12) Mass = moles xX molar mass 58.93 gCo 1kgCo ——— = 0. ImolCo * 1000gco~ °280k9 Co 13) Balance the chemical equation given below and determine the number of grams of MgO are needed to praduce 30.0 g of Fe203. MgO(s) + Fe(s) > Fe203(s) + Mg(s) 1 3) A)7.57g B) 2.529 C) 0.0440 g D) 22.7 g E) none of these 3MgO + 2Fe > Fe,0,+3Mg Mass of MgO 1mol Fe,0; _ 3molMgO 40.30 g MgO 30.0 g Fe,0. = 22.7 g MgO 5 F203” 759 70g Fe,0, 1molFe,0;1molMgO ams 14) How many moles of BCI3 are needed to produce 20.0 g of HCI(aq) in the following reaction? BCI3(g) + 3 H2O(!) > 3 HCl(aq) + B(OH)3(aq) 14) A) 0.549 mol B) 0.183 mol C) 1.65 mol D) 5.47 mol E) none of these 1mol HCl 1 mol BCI; 20.0gHCl* 3746 Gg HCL’ 3 mol HCl =.183 mol BCl3 15) What is the coefficient for oxygen when the following equation is balanced using the lowest, whole-number coefficients? —__ ©3Hg9 (I) + O2(9) > co2(g) + H20(!) 15) B 3 Balance reactants > Products c)5 3093 CO, - 9 ' 8H>4H.0 ion CsHg0 + __0, > 3C0, + 4H,0 Count oxygen in products: 10 Two oxygen comes from C3Hg0, which requires 10-1=9 more oxygen 26 32 9 CHg0 + 502 > 30, + 4H,0 2C3H,0 + 902 > 6CO, + 8H20 16) If 100. mL of 0.200 M Na2SO4 is added to 200. mL of 0.300 M NaCl, what is the concentration of Na* ions in the final solution? Assume that the volumes are additive. 16) a Ooo u Moles of Na,SQ4 = 0.200 M x 0.100 L = 0.0200 mol Cc) 0.267 M Moles of [Na*] from NazS0, = 2(0.0200mol) = 0.0400 mol D) 0.333 M Moles of NaCl = 0.300M x 0.200 L = 0.0600 mol E) none of these Moles of [Na*] from NaCl = 0.0600 mol Total moles of [Na*] = 0.0400 mol + 0.0600 mol = 0.1000 mol Total volume of solution: 0.300 L 0.1000 mol Concentration of [Na*] = ———_—— * 0.333M 0.300L 
