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

20L Cl2 * 1mol/22.4L = 0.89 mol
0.89mol Cl2 * 2 = 1.79 mole NaCl
1.79 mol NaCl * 58.44g = 104.6g

104.6g

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