[oceanmd]

Please check the problem. Thank you.

If 36.5 g of HCl and 73.0 g of Zn are placed together, and the mentioned reaction takes place: 2HCl + Zn = ZnCl2 + H2
Determine which reactant is limiting; determine which reactant is in excess and by how much.
Solution
36.5g HCl x (1 mol/1.01+35.45)=1.oo mol HCl
73g Zn x (1 mol/65.39)=1.116 mol Zn
2 mol of HCl are needed to react with 1 mol Zn
1.00 mol HCl/1.116 mol Zn = 0.87 mol HCl/1 mol Zn
Only 0.87 mol of HCl is available for every 1 mol of Zn instead of 2.
HCl is limiting
Zn is in excess

1 mol HCl is available, so only 0.5 mol of Zn reacts, because mole ratio is 2 mol HCl/1mol Zn
1.116mol Zn-0.5mol Zn=0.616 mol Zn is in excess
0.616 mol Zn x 65.39 g Zn=40.29g Zn in excess

[AWK]

HCl is limiting reagent - it is OK
But I would say I need 2.23 moles of HCl insteat of 1 mole to complete this reaction

[oceanmd]

AWK,
I agree that you need 2.23 mol of 2.23 to use up completely 73 g of Zn, but this is not is asked in the problem. It is asked by how much Zn is in excess. Zn is in excess by 40.29 g. Is this correct?

Thank you

[Sev]

Yes.

[AWK]

3 significant digits - 40.3 g