Question #256659

V（s） + O2G -V2O3（S）50.94g/mol + 32.00g/mol^ 149.88G/mol A.）计算钒（LLL）氧化物的理论产率，假设您以200.00克钒金属开头。B.）进行实验后，产生了183.2克的实验收率。计算该实验的百分比。C.）确定百分比误差。D.）计算过量反应物的过量量。

Expert's answer

The balanced equation:

4V(s) + 3O_{2}v (g)——> 2_{2}O_{3}(s)

A)

"Theoretical\\ yield\\ (V_2O_3)=200.00\\ g(V)\\times\\frac{1\\ mol(V)}{50.94\\ g(V)}\\times\\frac{2\\ mol(V_2O_3)}{4\\ mol(V)}\\times\\frac{149.88\\ g(V_2O_3)}{1\\ mol(V_2O_3)}=294.2\\ g"

**Answer:**294.2 g

B)

"\\%\\ yield=\\frac{actual\\ yield}{theoretical\\ yield}\\times100\\%=\\frac{183.2\\ g}{294.2\\ g}\\times100\\%=62.27\\%"

**Answer:**62.27 %

C)

"\\%\\ error=\\frac{|actual\\ yield-theoretical\\ yield|}{|theoretical\\ yield|}\\times100\\%=\\frac{|183.2\\ g-294.2\\ g|}{294.2\\ g}\\times100\\%=37.73\\%"

**Answer:**37.73 %

D)

The excess reactant leftover can only be calculated in case of assumption that the actual yield of the experiment is less than the theoretical yield only due to the shortage of one of the reactants (in this case - O_{2}because the amount of V is given), but for no other reason. Therefore, the amount of vanadium metal involved in the reaction can be found:

"m(V)=183.2\\ g(V_2O_3)\\times\\frac{1\\ mol(V_2O_3)}{149.88\\ g(V_2O_3)}\\times\\frac{4\\ mol(V)}{2\\ mol(V_2O_3)}\\times\\frac{50.94\\ g(V)}{1\\ mol(V)}=124.5\\ g"

So the mass of the excess reactant equals:

"m(V,excess)=200.00\\ g-124.5\\ g=75.5\\ g"

**Answer:**75.5 g

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