posted by noell on .
if you have one mole of CH3CH3(g) + 2 moles of Cl2 (g) and form one mole of C2H4Cl2 (l) + 2 moles of HCl (g)
assuming a 90% yield calculate the amount of each reactant needed to form 1.5 kg of product
I started the problem following the rules from the chemistry stochiometry section on the Jiskha site and am still confused when it comes to the product are they talking about C2H4Cl2 and HCl together and if they are together as the product do I combine their mole ratio and molecular weights in order to find the individual amounts of each reactant
I agree that the problem is not worded very well but I would interpret it as meaning that you want the two products together to be 1.5 kg after a 90% yield.
I would do this first---check my thinking since this is not the usual way of stating stoichiometry problems.
Since we know that 1 mole of C2H6 + 2 moles Cl2 will provide 1 mole C2H4Cl2 and 2 moles HCl, we ask ourselves what mass is 1 mole C2H4Cl2 and 2 moles HCl.
1 mole C2H4Cl2 = 98.95 g and 2 moles HCl = 72.92 g(but confirm those). The total is 171.87 so percent C2H4Cl2 is (98.95/171.87)*100 = about 57% or so and percent HCl is about 43% or so. If we want the total to be 1.5 kg (after a 90% yield), we must make the reaction think we want 1.5/0.9 = 1.67 kg. Then 1.67 x 57% = ?? and 1.67 x 43% = ??. Then you start with EITHER of those values (the ?? values) and work as the stoichiometry problem you have looked at to calculate kg CH3CH3 to start and kg Cl2 to start. If you wish to check yourself you may use the OTHER ?? value and recalculate everything but the amount CH3CH3 and Cl2 to start should be the same no matter which way you go.
Please post again with your work if you get stuck. I shall be happy to help you through.