Question

If 75 g of FeCO3 is heated with excess Oxygen, 45 g of Fe2O3 is produced. what is the percent yeild?????

Answer 1

ok first lets balance the chemical equation

Answer 2

we only need to worry about this sub reaction \[\text{FeCO}_3\to \text{Fe}_2\text O_3\]

Answer 3

its 4FeCO3 + O2 --->2Fe2O3 + 4CO2 thats what the book gives me

Answer 4

\[ 4\text{FeCO}_3\to 2\text{Fe}_2\text O_3\]
ok so the thing to notice is ratio of two
\[\color{brown} 2\text{FeCO}_3\to \text{Fe}_2\text O_3\]

Answer 5

Two moles of reactant will theoretically form one mole of Product

Answer 6

Can you find how many moles of reactant 75 g of FeCO_3 is ?

Answer 7

\[\boxed{ n=\dfrac mM}\]
\(n\) is number of moles \([\text{moles}]\)
\(m\) is mass \([\text g]\)
\(M\) is molar mass \(\frac{[\text g]}{[\text{moles}]}\)

Answer 8

\[M_{\text{FeCO}_3}=M_{\text{Fe}}+M_{\text{C}}+3M_{\text{O}}\]

Answer 9

\[M_{\text{Fe}}=55.85\\M_{\text{C}}=12.01\\M_{\text{O}}=16.00\]

Answer 10

are you following? @needsomuchhelp

Answer 11

the relative percentage yield is \[\frac{\text{Actual yield}}{\text{Theoretic yield}}\times100\%\]