For the reaction 2na(s) + cl2(g) --> 2nacl(s), how many grams of nacl could be produced from 20 l of cl2 (at stp)?

13.0 L Cl2 × 1 mol Cl2 /22.4 L Cl2 = 0.580 mol Cl2 103.0 g Na × 1 mol Na/23 g Na = 4.48 mol Na Cl 2 is limiting reagent: 0.580 mol Cl2 × 2 mol NaCl/1 mol Cl2 = 1.16 mol NaCl 1.16 mol NaCl × 58 g NaCl/1 mol NaCl = 67.3 g NaCl DIF: L3 REF: p. 371 OBJ: 12.3.1

B. a graduated substance