Hi everyone, I have this problem that I'm somewhat confused about. The problem is:An alloy of aluminum and magensium was treated with sodium hydroxide solution, in which only aluminum reacts to give hydrogen gas:2Al + 2NaOH + 6H20 --gt; 2NaAl(OH)4 + 3H2If a sample of alloy weighing 1.118 g gave 0.1068 g of hydrogen, what is the percentage aluminum in the alloy?How do I account for the magnesium and what stoichiometric relationships would I need to set-up? This is very confusing, but I appreciate the slightest of help on this!Thank you.
You don't need to worry about the magnesium at all, because it doesn't enter into the reaction. You have a balanced equation for aluminium and hydrogen, and the data given are sufficient: From the equation, you know that 2 moles of aluminium produce 3 moles of H2. 0.1068 g of H2 were produced, so divide this by molecular mass of H2 (2.016) to determine the moles of H2 produced. You know that 3 moles of H2 would have been produced from 2 moles of Al. Therefore, multiply the moles of H2 by 2/3, to determine the moles of Al that were present. Now multiply this number by the atomic mass of Al, to determine the grams of Al. Now divide this number if grams by 1.118 and multiply by 100 to determine % Al present in the alloy.
