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Answer to Question #73496 in General Chemistry for Tazyika Markman
Question #73496
combining 0.385 mol of feo3 with excess carbon produced 16.5g of fe.
what is the actual yield of iron in moles?
what was the theoretical yield of iron in moles?
what was the percent yield?
what is the actual yield of iron in moles?
what was the theoretical yield of iron in moles?
what was the percent yield?
Expert's answer
.n (FeO3)=0.385 mol
.m (Fe)=16.5 g
.n(Fe)theor=?
%=?
FeO3+ 3C=Fe+3CO
.n (Fe)theor= n (FeO3)=0.385 mol
Amount of substance of iron n (Fe)=m/Ar= 16.5/56=0.295 mol
%= n(Fe)/ n (Fe) theor= 0.295/0.385 *100 %= 76.62 %
.m (Fe)=16.5 g
.n(Fe)theor=?
%=?
FeO3+ 3C=Fe+3CO
.n (Fe)theor= n (FeO3)=0.385 mol
Amount of substance of iron n (Fe)=m/Ar= 16.5/56=0.295 mol
%= n(Fe)/ n (Fe) theor= 0.295/0.385 *100 %= 76.62 %
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The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.876 g sample of BHT was combusted in an oxygen rich environment to produce 11.61 g of CO2(g) and 3.803 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT.
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