If you think to the reaction:
Al4C3 + 12 H2) = 4 Al(OH)3 + 3 CH4
the mass of aluminium hydroxide is 0,746 g.
To find the number of moles of Al(OH)3 produced, we need to balance the chemical equation for the reaction where CH4 is formed. Once the equation is balanced, we can determine the mole ratio between Al(OH)3 and CH4. If we assume the balanced equation is: 4CH4 + 3O2 -> 2Al(OH)3 + 4H2O, we can see that the mole ratio is 2:4. Therefore, for every 2 moles of Al(OH)3 produced, 4 moles of CH4 are needed. Since we have 0.560 mol of CH4, we will produce half as many moles of Al(OH)3, which is 0.280 mol. To find the mass in grams, we can multiply the moles by the molar mass of Al(OH)3, which is 78 g/mol. Therefore, 0.280 mol of Al(OH)3 will be produced, which is equal to 21.84 grams.
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