__ Al(s) + __ HCl(aq) --> __ AlCl3(aq) + __H2(g)

[Chem_Mod]

Aluminum and hydrogen chloride, HCl, react to form aluminum chloride in solution, AlCl3, and hydrogen gas. Balance the equation. __ Al(s) + __ HCl(aq) --> __ AlCl3(aq) + __H2(g) If you had 5.43 g of aluminum and 7.80 g of HCl how many grams of AlCl3 would you get? I have the balanced equation but I can't solve the second part of the question.

[Chem_Mod]

For the second part of this equation, you need to find the limiting reactant. You first determine how much AlCl3 would be produced with 5.43 g or Al by converting g Al into mol Al, then multiplying by the molar ratio of AlCl3 to Al (mol AlCl3/mol Al from the balanced equation). This will give you the number of mols of AlCl3 produced from Al. Then you do the same calculation for HCl. Whichever reactant would produce the least amount of product is your limiting reactant. Now that you know the limiting reactant, you use it to determine the amount of product that can be formed. You already calculated the amount of mol of product produced, so just multiply by the molar mass of AlCl3 and that is g of AlCl3 you would get.

[DanielaWong_2G]

I did that, but i got a different answer than the one in the book. I got 4.49 but it says the answer is 9.51 and i can't figure out what I did wrong

[Logan 3C]

Make sure you base your calculations on how many moles of an element in the molecule because that can really change your answer. For example if the molecule is AlCl3, make sure to calculate its molecular mass using 3 Cl atoms.