What volume of chlorine is required for the reaction with sodium of 34.5 g to calculate the mass of sodium chloride?

univerkov | December 8, 2020 | education

2Na + Cl2 = 2NaCl
m (Na) = 34.5 grams
n (Na) = m / M = 34.5 / 23 = 1.5 mol
the equation shows that chlorine is needed 2 times less than 0.75 mol
V (Cl) = n * Vm = 0.75 * 22.4 = 16.8 liters of chlorine
the equation shows that n (Na) = n (NaCl) = 1.5 mol of sodium chloride
m (NaCl) = n * M = 58.5 * 1.5 = 87.75 g.
Answer: V = 16.8 liters; m = 87.75

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