Question

When 7.0 mol Al react with 8.5 mol HCl, what is the limiting reactant and how many moles of AlCl3 can be formed?

2 Al + 6 HCl “yields”/ 2 AlCl3 + 3 H2

Al is the limiting reactant, 7.0 mol AlCl3 can be formed

HCl is the limiting reactant, 2.8 mol AlCl3 can be formed

Al is the limiting reactant, 3.5 mol AlCl3 can be formed

HCl is the limiting reactant, 8.5 mol AlCl3 can be formed

I picked the last one. Is it right?

Answer 1

my work was 7.0 mol Al * 3 mol H2/2 mol Al * 2.016 g H2/1 mol H2 = 21.168 g H2

then 8.5 mol HCl * 3 mol H2/6 mol HCl * 2.016 g H2/1 mol H2 = 8.568 g H2

Answer 2

please someone help.

Answer 3

you're right about HCl being the limiting reactant but the amount of moles of AlCl3 isn't correct the correct answer is 2

Answer 4

@amematsro can you explain?

Answer 5

like what would be the answer then? b?

Answer 6

okay so you have 8.5 moles of HCl and for every 6 mols of HCl you can make 2 moles of AlCl3, so what you want to do is take 8.5 the number of moles of HCl that you have and divide it by 6 the number of moles you need to have then you multiply that by 2 because you make 2 moles of AlCl3, get it?

Answer 7

ohhh ok. So the answer is B, correct?

Answer 8

yup

Answer 9

thanks so much man. Youre a life saver. Could you possibly answer another question for me? This is way shorter

Answer 10

Yeah sure no problem

Answer 11

Could you balance this equation for me? Fe + AgNO3 “yields”/ Fe(NO3)3 + Ag

Answer 12

It's Fe + 3 AGNO3 = Fe(NO3)3 + 3 AG right?

Answer 13

yeah that is perfect

Answer 14

Thank you! Is it okay if I ask you questions later on? I mean, I'm okay right now but is it fine if I hit you up later?

Answer 15

@amematsro