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For the reaction 2Na(s) + Cl2(g) -> 2NaCl(s), how many grams of NaCl could be produced from 20 L of Cl2 (at STP)?
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For the reaction 2Na(s) + Cl2(g) -> 2NaCl(s), how many grams of NaCl could be produced from 20 L of Cl2 (at STP)?

User Jacob Barnes
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1 Answer

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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
User David Vrba
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