How many liters are released during the hydrolysis of one mole of aluminum carbide

univerkov | December 3, 2020 | education

Al4C3 + 12H2O = 4Al (OH) 3 + 3CH4
n (Al4C3) = 1 mol according to the condition
n (CH4) according to the reaction equation = 3 mol
V (CH4) = n * Vm = 3 * 22.4 = 67.2 l
Answer: 67.2

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