Question

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

Answer 1

4V + 3O₂ → 2V₂O₃

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

According to the reaction:

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

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

Answer: 294.29 g