Is Anyone great at Chemistry

<p>I am taking an AP chem course with a terrible professor, he assigned us a stickyometry problem( in this case my nick name for these nasty little problems fits perfectly). The problem is as follows:</p>

<p><em>Zn + _HCl yields _ZnCl</em> + _H2</p>

<p>the assignment associated was to find the amount of moles of Hydrogen based on the gas law pv=nrt...I did that and I got 1.7 moles, given was .0017 moles of Zn were used...I have to find the coefficients and the subscript for Cl...If anyone can help me I would be so grateful...I have an idea of how to continue the problem, but I'm just not sure...Thank you for your help...</p>

<p>please help me</p>


<p>Shouldn't you be trying to do your own work u'll learn more</p>

<p>The balance equation is Zn + 2HCL --> ZnCl2 + H2
but i don't see the point in finding the moles of H2 using the PV=nRT formula. Did you have any given information about the pressure and the temperature? You should know just which one is the limiting reactant. I guess in your case it's Zn. The moles of Zn to the moles of H2 are as 1:1, so there are .0017 moles H2.</p>

<p>The information given included temperature, atm pressure, and volume and originally the only information given was .0017 moles of Zn as the limiting reagent...The problem is I have to show how to mathmatically obtain the subscript of ZnCl because the next question is to use the same method and the same information to find the subscripts in SnCl if Sn is put in place of Zn,,,I know what the answer is I just don't know how to get from the info I have now to that answer</p>

<p>sorry, cant help u :o)</p>

<p>Thanks anyway</p>

<p>You have to use their oxidation(?) numbers, which is the superscript. There's no way I can think of to truly do it "mathematically." Transition metals are usually a +2 charge, and the other non-transition elements are simple to get.</p>

<p>For the coefficients, just balance it as someone did above.</p>

<p>You should give the entire problem and all the information to get the best help.</p>