What is the volume and mass of hydrogen released during the interaction of 6.9 g of sodium metal with water?

univerkov | March 18, 2021 | education

2Na + 2H2O = 2NaOH + H2
n = m / M (Ar) M = Ar Ar (Na) = 23
n (Na) = 6.9 / 23 = 0.3 (mol)
n (Na) / n (H2) = 2/1, next, n (H2) = 0.3 / 2 = 0.15 (mol)
m = n * M (Mr) M = Mr Mr (H2) = 2 M = 2g / mol
m (H2) = 0.15 * 2 = 0.3 (g)
V = n * Vm V (H2) = 0.15 * 22.4 = 3.36 (l)
Answer: 3.36l, 0.3g

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