this class is killing me

how many grams of aluminum chloride is being produced when 1.52 x 10^24 atoms of aluminum react with sufficient amount of chlorine gas?

how many grams of c3h3n can be produced when 25.6g of c3h6 reacts with 23.5g of nitric oxide. determine the limiting reagent as part of the answer

if anyone can do these and show some work would be greatly appreciated

Divide that number by avogados number to obtain the mols of Al.

Every mil of AlCl3 has one Al, and you have all the Cl you need, however much Al you got will be the answer, because it can all react.

Write down the chemical reaction listed. Figure out the coefficients. Convert grams to mols by diving my molar mass.

Chemistry?

You and me got chemistry bae, Im gonna fuggin H2 your O

first question:
how many grams of aluminum chloride is being produced when 1.52 x 10^24 atoms of aluminum react with sufficient amount of chlorine gas?

first use stoichiometry to find aluminum and chlorine react in this way: Al + 3Cl <---> AlCl3 [3 should be subscript]
now realize that aluminum is the limiting reagent (they tell us we get sufficient chlorine). Thus, for every Al atom, we get one Aluminum Chloride molecule.
Thus we will have 1.52 x 10^24 molecules. Now find this molecule's mass in g/mol on the periodic table by combining the required atoms.
Al = 26.982g/mol
Cl = 35.453g/mol

26.982 + 3(35.453) = 133.341 g/mol.

So how many moles do we have? 1 mole = 6.02x10^23, so we just divide our number of molecules by Avagadro's number giving us
(1.52 x 10^24) / (6.02 x 10^23) = 2.52 moles
now multiply 2.52mol x 133.341g/mol = 336.67 grams

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