Chemistry Question! Confused!!!?

Question

What mass, in grams, of ZnCl2 can be prepared from the reaction of 3.27 grams of zinc with 3.30 grams of HCl? Zn + 2HCl → ZnCl2 + H2

Answer 1

Mike, you have to convert your mass info to moles, use the balanced reaction to find the moles of ZnCl2 per mole of HCl, and then convert back to mass for ZnCl2 (moles*mole wt). ALSO, you have to determine which reactant is limiting, the zinc or the HCl.

Approximately, HCl moles are 3.30/38= 0.085
From reaction, we have enough Cl- for 0.042 moles of ZnCl2.
0.042 mole of ZnCl2 is appx 0.042*125 g/mole= 5 grams ZnCl2.
Do we have enough Zinc? 3.27 g/55g/g-mole=0.06 moles. Since we can only react with 0.042 moles Zn, we have enough.

Answer 2

for this, you need to take into account the atomic mass of each item, and the relative contribution of each to make the zinc dichloride.

3.27g Zn @ 65.4g/mol = 0.05mol Zn
3.3g HCL @ 36.461 (35.453+1.008)g/mol = 0.0905 mol Cl

now, to make ZnCl2, we need twice as much Cl as Zn. Thus, if we used all of the Zn, we would need 0.1mol of Cl. But we don't have that much! We only have 0.09. Therefore, we use Cl as the "limiting reagent" and use that as the calculation. Always remember: whatever reagent you have the least of, that is what you will calculate your product from.

So, with for every 1 mol of Cl, we make 1/2 mol of ZnCl2.

0.0905/2 = 0.04525 mol ZnCl2 @ 136.31g/mol = 6.17g ZnCl2