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Question
Aluminum carbide reacts with water according to the following equation:
\[\ce{Al4C3 + 12H2O -> 4Al(OH)3 + 3CH4}\]
- What mass of aluminum hydroxide is formed from 12g of aluminum carbide?
- What volume of methane at s.t.p. is obtained from 12g of aluminum carbide?
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Solution
a. \[\ce{\underset{144 g}{Al4C3} + 12H2O -> \underset{312 g}{4Al(OH)3} + \underset{67.2 lit}{3CH4}}\]
144 g of aluminium carbide forms 312 g of aluminium hydroxide.
12 g of aluminium carbide will form:
`312/144 xx 12` = 26 of aluminium hydroxide
Hence, 26 g of aluminium hydroxide is formed.
b. 144 g of aluminium carbide forms 67.2 lit of methane.
∴ 12 g of aluminium carbide will form:
`67.2/144 xx 12` = 5.6 lit.
Hence, the volume of methane obtained at s.t.p. is 5.6 lit.
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