# chemistry

posted by .

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

• chemistry -

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.

## Respond to this Question

 First Name School Subject Your Answer

## Similar Questions

1. ### chemistry

Equation- Mg + 2HCl -> MgCl2 + H2 What volume of hydrogen at STP is produced from the reaction of 50.0 g of Mg and the equivalent of 75g of HCl?
2. ### Chemistry

When 9.0g of Al were treated with an excess of chlorine, 20.0g of Al2Cl6 were collected. 2AL + 3Cl6 -------> Al2Cl6 What was the percentage yield?
3. ### Organic Chem/ Calculate Yield

How would you calculate the percent yield for a reaction that began with 10 moles of acetone and 16 moles of benzaldehyde which yielded 6 moles of dibenzalacetone as the product?
4. ### Chemistry 130

Will someone help me by letting me know if I did these correct, or if there the correct answers?
5. ### chemistry

given the reactant amounts specified in each chemical eguation, determine the limiting reactant in each case: a. HCL+NaOH->NaCl+H2O 2.0 mole of HCl 2.5 mole NaOH b. Zn+2HCl->ZnCl2+H2 2.5 mole Zn 6.o mole HCl c. 2Fe(OH)3+3H2SO4->Fe2(SO4)3+6H2O …
6. ### chemistry

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
7. ### Chemistry

A mixture of 0.47 mole of H2 and 3.59 moles of HCl is heated to 2800C. Calculate the equilibrium partial pressures of H2 Cl2 and HCl if the total pressure is 2.00 atm. For the reaction Kp is 193 at 2800C. H2(g) +Cl2(g) = 2HCl (g) I …
8. ### Chemistry

A mixture of 0.47 mole of H2 and 3.59 moles of HCl is heated to 2800C. Calculate the equilibrium partial pressures of H2 Cl2 and HCl if the total pressure is 2.00 atm. For the reaction Kp is 193 at 2800C. H2(g) +Cl2(g) = 2HCl (g) What …
9. ### chemistry

Record and calculate the following masses: (a)mass of empty beaker: 47.420g (b)mass of beaker plus Na₂CO₃:47.920g (c)Mass of Na₂CO₃:0.5g (d)mass of beaker plus NaCl:47.438g (e) Mass of NaCl (g):0.018g **I need …
10. ### chemistry

What is the mass of 4.80 moles of MgCl2?My answer is: Mg=24.3 g/mole Cl2=35.5g/mole*2=71g/mole so, Mg 24.3+ Cl2 71= 95.3g/mole therefore 4.80 moles * 95.3 g/mole=471.84g Am i right pls correct me if i'm wrong thanks

More Similar Questions