Asked by Kami
Question
Given the equation: 4 Fe(s) + 3 O2(g) ? 2 Fe2O3(s), what is the theoretical yield of Fe2O3 from the reaction of 4.86 g Fe with excess O2? b. If 6.76 g Fe2O3 are actually obtained from 4.86 g Fe, what is the percent yield? c. Under other conditions, the percent yield is 75.6%. How many grams of Fe2O3 are formed from 4.86 g Fe under these conditions?
Given the equation: 4 Fe(s) + 3 O2(g) ? 2 Fe2O3(s), what is the theoretical yield of Fe2O3 from the reaction of 4.86 g Fe with excess O2? b. If 6.76 g Fe2O3 are actually obtained from 4.86 g Fe, what is the percent yield? c. Under other conditions, the percent yield is 75.6%. How many grams of Fe2O3 are formed from 4.86 g Fe under these conditions?
Answers
Answered by
DrBob222
mols Fe = 4.86 g Fe/atomic mass Fe = ?
Using the coefficients in the balcned equation, convert mols Fe to mols Fe2O3.
Now convert mols Fe2O3 to grams Fe2O3. g = mols x molar mass = ? and this is the theoretical yield (TY) The actual yield (AY) is 6.76.
%yield = (AY/TY)*100 = ?
Just rearrange the %yield equation above from
%yield = (AY/TY)*100
75.6 = (AY/TY)*100 and solve for AY.
Using the coefficients in the balcned equation, convert mols Fe to mols Fe2O3.
Now convert mols Fe2O3 to grams Fe2O3. g = mols x molar mass = ? and this is the theoretical yield (TY) The actual yield (AY) is 6.76.
%yield = (AY/TY)*100 = ?
Just rearrange the %yield equation above from
%yield = (AY/TY)*100
75.6 = (AY/TY)*100 and solve for AY.
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