Answer to Question #256424 in General Chemistry for Jonalyn Galao Gamu

Question #256424
V(s)50.94g/mol+O2g32.00g/mol----V2O3(s)149.88g/mol calculate the theoretical yield of vanadium (lll) oxide, assuming you begin with 200.00 grams vanadium metal
1
Expert's answer
2021-10-26T05:34:07-0400

4V + 3O2 → 2V2O3

m(V) =200.050.94=3.926  mol= \frac{200.0}{50.94} = 3.926 \;mol

According to the reaction:

n(V2O3) =12n(V)=12×3.926=1.963  mol= \frac{1}{2}n(V) = \frac{1}{2} \times 3.926 = 1.963 \;mol

m(V2O3) =1.963×149.88=294.29  g= 1.963 \times 149.88 = 294.29 \;g

Answer: 294.29 g


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog