Question

Give the percent yield when 28.16 g of co2 are formed from the reaction of 8.000 moles of c8h18 with 4.000 moles of o2. 2 c8h18 + 25 o2 → 16 co2 + 18 h2o

Answer 1

equal moles of C8H18 and O2 are reacted to give equal no. of moles of CO2.

mole ratio of C8H18 =(1/2) x 8 = 4 moles

mole ratio of O2 = (1/25) x 4 = 0.16 moles

so, limiting reagent is O2

the no. of moles of CO2 formed = 16 x 0.16 = 2.56 moles

weight of CO2 formed (theoretical weight) = 2.56 x 44 = 112.64

Percentage yield =(practical yiel/theoretical yield) x 100 = (28.16/112.64) x100 = 25%

Answer 2

Answer: The percent yield of carbon-dioxide gas is 25%.

Step-by-step explanation:

`2C_8H_(18)+25O_2\rightarrow 16CO_2+18H_2O`

1) According to reaction 25 moles of oxygen gas reacts with 2 moles of
`C_8H_(18)`, then 4 moles of oxygen will react with
`(2)/(25)* 4` moles of
`C_8H_(18)` that is 0.32 moles.

Oxygen with 4 moles is limiting reagent in this reaction.

2) According to reaction 25 moles of oxygen gas gives 16 moles of carbon-dioxide gas, then 4 moles of oxygen gas will give
`(16)/(25)* 4` moles of carbon-dioxide gas that is 2.56 moles.

Theoretical mass of
`CO_2=\text{number of moles}* \text{molecular mass of }CO_2}=2.56mol* 44g/mol=112.64 g`

Experimental mass of
`CO_2` = 28.16 g

`Percent yield=\frac{|\text{Experimental mass}}{\text{Theoretical mass}}=(28.16* 100)/(112.64)=25\%`

Hence, the percent yield of carbon-dioxide gas is 25%.