What volume of hydrogen will be released when 55.2 g of ethyl alcohol interacts with 29.9 g of metallic sodium?

univerkov | July 25, 2021 | education

2C2H5OH +2 Na = 2C2H5ONa + H2.

M (C2H5OH) = 2 * 12 + 5 * 6 + 16 = 70 g / mol.

n (C2H5OH) = 55.2 / 70 = 0.79 mol.

M (Na) = 23 g / mol.

n (Na) = 29.9 / 23 = 1.3 mol.

X = 2 * 1.3: 2 = 1.3 mol.

Therefore, sodium was taken in excess. We count on alcohol.

Y = 1 * 0.79: 2 = 0.395 mol.

V (H2) = 22.4 * 0.395 = 8.848 HP

8.848 liters of hydrogen will be released.

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