Question

44.00g/mol 4. RbBr 0 RbBr Use the following information to answer problems 11 and 12. 0.462 moles of CS, and 0.910 moles of Oz gas react according to the balanced chemical equation 0.462 moles Cs2 Imolen ICS2(1) + 302(g) → CO2(g) + 250 (9) 11. What is the limiting reactant? The molar mass of CO2 is 44.00 g/mol. 20.0105 1. CS, is the limiting reactant and 20.3 g of CO2 are formed 0.910 mo 02 2. O2 is the limiting reactant and 20.3 g of CO2 are formed. moloz 3. CS2 is the limiting reactant and 13.3 g of CO2 are formed. 4. O2 is the limiting reactant and 13.3 g of CO2 are formed. 0105 0.028. 12. What amount of excess reactant is present after the reaction is finished? The molar mass of 0.018 CS2 is 76.14 g/mol and O2 is 32.00 g/mol. 3 Total-reactant-lef over 1. 12.1 g remain of the excess reactant CS2 nnn n nnm -0.009 2. 6.1 g remain of the excess reactant CS2 merenowym makes more 3. 15.3 g remain of the excess reactant O2 4 2.9 g remain of the excess reactant O2 0.010s CC 13. What is the oxidation number of chlorine in sodium perchlorate, NaClO.? Erint

Answer 1

no needed=3 molo, x 0.462 molch Inocsa = 1.386 mola but we have only 0.91 molo o is limiting reactant nco, forned = 1 nolco₂ x 0.91molo2 3 nolo2 0.303 md. mco, forned=0.303 mox 44 g/mol = 13.39 Co2 Nese left = 0.46 2. cnol – 0.-303 -0.159 no mcs. left = 0.159 and x 76 luglmo = 12-19 Naclou: cl- oxidation number is "+7