What volume of gas will be released when 90 g of technical lithium containing 10% impurities is dissolved in water?

univerkov | December 17, 2020 | education

To begin with, in solving this problem it is necessary to compose the reaction equation:
2Li + 2H2O = 2LiOH + H2 (gas)
m (Li) = 90g * 0.1 = 9g (mass of lithium)
n (Li) = 9g / 7g / mol = 1.3 mol (amount of substance)
V (Li) = 1.3 mol * 22.4 l / mol = 29 l (volume)
Answer: V (Li) = 29 l

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