Answer :
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: The percent yield of carbon-dioxide gas is 25%.
Explanation:
[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]
1) According to reaction 25 moles of oxygen gas reacts with 2 moles of [tex]C_8H_{18}[/tex], then 4 moles of oxygen will react with [tex]\frac{2}{25}\times 4[/tex] moles of [tex]C_8H_{18}[/tex] 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 [tex]\frac{16}{25}\times 4[/tex] moles of carbon-dioxide gas that is 2.56 moles.
Theoretical mass of [tex]CO_2=\text{number of moles}\times \text{molecular mass of }CO_2}=2.56mol\times 44g/mol=112.64 g[/tex]
Experimental mass of [tex]CO_2[/tex] = 28.16 g
[tex]Percent yield=\frac{|\text{Experimental mass}}{\text{Theoretical mass}}=\frac{28.16\times 100}{112.64}=25\%[/tex]
Hence, the percent yield of carbon-dioxide gas is 25%.