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.

To solve this problem, you need to follow the steps of stoichiometry to determine the amounts of each reactant needed to form 1.5 kg of product.

Step 1: Write the balanced chemical equation:
CH3CH3(g) + 2Cl2(g) → C2H4Cl2(l) + 2HCl(g)

Step 2: Convert the given mass of the product into moles:
1.5 kg of C2H4Cl2 = 1500 g
Molar mass of C2H4Cl2 = 98.96 g/mol (C = 12.01 g/mol, H = 1.01 g/mol, Cl = 35.45 g/mol)

Number of moles of C2H4Cl2 = (1500 g) / (98.96 g/mol) = 15.15 mol (rounded to two decimal places)

Step 3: Determine the reactant ratios from the balanced chemical equation:
From the balanced equation, 1 mol of CH3CH3 reacts with 1 mol of C2H4Cl2 produced.
Therefore, the mole ratio of CH3CH3 to C2H4Cl2 is 1:1.

Step 4: Calculate the moles of CH3CH3 needed:
Number of moles of CH3CH3 = 15.15 mol

Step 5: Calculate the moles of Cl2 needed:
From the balanced equation, 2 mol of Cl2 reacts with 1 mol of C2H4Cl2 produced.
Therefore, the mole ratio of Cl2 to C2H4Cl2 is 2:1.

Number of moles of Cl2 = (15.15 mol) * (2/1) = 30.30 mol (rounded to two decimal places)

Step 6: Adjust for the given 90% yield:
Since the given yield is 90%, we need to adjust the calculated amounts of CH3CH3 and Cl2 according to their stoichiometric ratios. We assume that the limiting reactant is CH3CH3 (the reactant that gets consumed first).

Adjusted moles of CH3CH3 = (15.15 mol) * (90/100) = 13.64 mol (rounded to two decimal places)

Adjusted moles of Cl2 = (30.30 mol) * (90/100) = 27.27 mol (rounded to two decimal places)

Step 7: Convert the adjusted moles back to grams:
Mass of CH3CH3 = (13.64 mol) * (30.07 g/mol) = 410.97 g (rounded to two decimal places)

Mass of Cl2 = (27.27 mol) * (70.90 g/mol) = 1937.17 g (rounded to two decimal places)

Therefore, to form 1.5 kg of C2H4Cl2 with a 90% yield, you would need approximately 411 grams of CH3CH3 and 1937 grams of Cl2.