What volume of gas will be released during the interaction of 11.7 g of sodium chloride with 44.8 liters of fluorine?

univerkov | December 17, 2020 | education

First, to solve this problem, you need to write the reaction equation:
2NaCl + F2 + 2NaF + Cl2
n (NaCl) = 11.7 / 58.5 = 0.2 mol (amount of substance) is in short supply, so we take this value and calculate it.
n (F2) = 44.8 / 22.4 = 2 mol.
n (Cl2) = n (NaCl) /2=0.2/2 = 0.1 mol
V (Cl2) = 0.1 * 22.4 = 2.24 l
Answer: V (Cl2) = 2.24 l

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