A mass of 8.15g C^2H^4(g) reacts with excess oxygen. If 16.2g CO^2(g( is collected, what is the percent yield?

1. Write down the equation for the reaction taking place.

C_2H_4 (g) + 3O_2 (g) \rightarrow 2CO_2(g) + 2H_2O(g)

2. Find the number of moles of C2H4 in the 8.15 g

3. Find the expected amount of moles of CO2 to be produced.

4. Find the mass of that number of moles.

5. Use the formula \frac{Actual\ yield}{Expected\ yield} \times 100% = Percentage\ Yield

EDIT: Missed something :o

Ok I will try to give ans
first write down alance combustin reaction for given compoud
C2H4 +3O2==========> 2CO2+2H2O
28 gm 96 gm 88gm 32gm


28 gm C2H4 produce 88gm CO2
1gm produce 88/28 gm CO2
8.15 gm will produce 88*8.15/28 gm co2
whch is equal to 25.61 gm
but according to your experiment only 16.2 gm found so it mean uallow to scape some co2 from your cane
wel percentage yield is 16.2gm*100/25.61=63.26% I think
I am right please reply me
pramod