Consider the reaction of aluminum with hydrogen chloride to produce aluminum chloride and hydrogen gas. If you react 151g of aluminum with 592g of hydrogen chloride, how much hydrogen gas could be produced, in grams?
balanced equation.
The balanced equation is 2 Al + 6 HCl = 2 AlCl3 + 3 H2
The equal sign shud be a yield sign
I found the moles of A and moles of HCl but im not sure if that's necessary
i think the answer is 16.9 because the HCl is the limiting reactant right?
16.9g H2
write the balanced eqn. find the Limiting reactant.
convert hcl to moles then mole mole ratio then mass h2
Aluminum is the limiting reagent i think since its 0.55 mol and HCl is 16.24 mol
How did u get 16.9. like what were ur steps
16.4 sorry
yeah that is teh right answer b/c i have teh asnwer key but i dont understand how to get taht answer
well i determined the theoretical yield of h2 produced from the balanced eqn
and you to that by
taking 151g Al, convert to moles, (times)mole ratio of aluminum to h2, then multiply by the mass of h2
and that will give you the mass of h2 produced.
but you also have 592g HCl so u do the same exact thing except with HCl
THE SMALLER NUMBER IS THE THEORETICAL YIELD or the amount that can be produced by the reaction
theoretically
so i got 0.55 mol Al. Now I multiple it by (3 H2/ 2 Al) = 0.825 mol H2 and 16.24 mol HCl 8 (3 mol h2/ 6 mo HCl) = 8.12 mol HCl.
ok reset bro start with grams Al
i mean HCl is limiting reagent right? then u multiply 8.12 * 2 = 16.24 g
yeah
8.2 is the number of moles of hydrogen
ok just alittle confusion. Why is the larger number of moles the limiting reagent?
because mole mole ratio changes it
Oh I get it now! Thanks :)
yeah so for this problem start with grams, goto moles, and back to grams. also make sure you keep track of units
units are very important
youre welcome btw. take a look at the question i posted if u know where to start
lmk
ok sure
Join our real-time social learning platform and learn together with your friends!