posted by Jack on .
Please help with this problem =)
The combustion of liquid chloroethylene, C2H3Cl, yields carbon dioxide, steam, and hydrogen chloride gas.
I. write a balanced equation for the reaction (I sorta get this part)
II. How many moles of oxygen are required to react with 35.00g of chloroethylene?
III. If 25.00g of chloroethylene reacts with with an excess of oxygen, how many grams of each product are formed?
Thanks again =)
2C2H3Cl + 5O2 ==> 4CO2 + 2H2O + 2HCl
Is that what you obtained?
mols in 35.00 g chloroethylene = 35.00/molar mass = 35.00/62.499 = 0.5600 mols. Convert to mols O2.
0.5600 mols C2H3Cl x (5 mol O2/2 mol C2H3Cl) = ??
25.00 g C2H3Cl = ?? mols = 25.00/62.499 = 0.4000 mols C2H3Cl.
a. Convert to mols CO2.
0.4 mol C2H3Cl x (4 mols CO2/2 mol C2H3Cl) = 0.8000 mol CO2.
g CO2 = mols CO2 x molar mass CO2.
b and c are done the same way. Note that you start with what you know (in this case mols C2H3Cl) and multiply by the coefficients in the balanced equation to convert mols C2H3Cl to any of the other materials. How do you know which coefficient is on top and which on bottom? Look at the units. You want the unit of C2H3Cl (in this case) to cancel and you want the unit CO2(in this case) to stay. Likewise, you want the H2O coefficient on top for the water conversion and HCl coefficient on top to convert to HCl.
Post your work if you get stuck.