Question

1. 1.00 x 10² g of aluminum and 1.50 x 10² g of oxygen are reacted according to the chemical equation below. 4al + 3o₂ → 2al₂o₃ a. what are the final masses of: i. the product ii. the reactants when the reaction has gone to completion? b. if 132.1 g of al₂o₃ is isolated after the reaction has been completed, what is the % yield of the alumina (al₂o₃)? c) h₂so₄(aq) + bacl₂(aq) → hcl(aq) + baso₄(s) h₂so₄ + bacl₂ → 2hcl + baso₄

Explanation:

Step1: Calculate molar masses

The molar mass of $Al$ is approximately $26.98\ g/mol$, of $O_2$ is $32.00\ g/mol$, and of $Al_2O_3$ is $2\times26.98 + 3\times16.00= 101.96\ g/mol$.

Step2: Determine moles of reactants

Moles of $Al$: $n_{Al}=\frac{1.00\times 10^{2}\ g}{26.98\ g/mol}\approx3.706\ mol$. Moles of $O_2$: $n_{O_2}=\frac{1.50\times 10^{2}\ g}{32.00\ g/mol}\approx4.688\ mol$.

Step3: Identify limiting reactant

From the balanced equation $4Al + 3O_2
ightarrow2Al_2O_3$, the mole - ratio of $Al$ to $O_2$ is $\frac{4}{3}$. For $n_{Al} = 3.706\ mol$, the moles of $O_2$ required is $n_{O_2 - required}=\frac{3}{4}\times3.706\ mol\approx2.7795\ mol$. Since the available $O_2$ is $4.688\ mol$, $Al$ is the limiting reactant.

Step4: Calculate theoretical yield of $Al_2O_3$

From the stoichiometry, for every 4 moles of $Al$, 2 moles of $Al_2O_3$ are formed. So, moles of $Al_2O_3$ formed $n_{Al_2O_3}=\frac{1}{2}\times n_{Al}=\frac{1}{2}\times3.706\ mol = 1.853\ mol$. The theoretical mass of $Al_2O_3$ is $m_{theoretical}=n_{Al_2O_3}\times M_{Al_2O_3}=1.853\ mol\times101.96\ g/mol\approx188.9\ g$.

Step5: Calculate percent yield

Percent yield $=\frac{actual\ yield}{theoretical\ yield}\times100\%=\frac{132.1\ g}{188.9\ g}\times100\%\approx69.9\%$

i. Theoretical mass of product ($Al_2O_3$) is approximately $188.9\ g$, actual mass is $132.1\ g$.
ii. After reaction completion, limiting reactant $Al$ is fully consumed, and there is excess $O_2$. Mass of un - reacted $O_2$: Initial moles of $O_2$ is $4.688\ mol$, moles of $O_2$ reacted $n_{O_2 - reacted}=\frac{3}{4}\times3.706\ mol = 2.7795\ mol$. Un - reacted moles of $O_2$ is $n_{O_2 - unreacted}=4.688\ mol - 2.7795\ mol = 1.9085\ mol$. Mass of un - reacted $O_2$ is $m_{O_2 - unreacted}=1.9085\ mol\times32.00\ g/mol\approx61.07\ g$. So mass of reactants: $Al$ is $0\ g$ (fully reacted), $O_2$ is approximately $61.07\ g$.

Answer:

i. Theoretical mass of $Al_2O_3$ is approximately $188.9\ g$, actual mass is $132.1\ g$.
ii. Mass of $Al$ is $0\ g$, mass of $O_2$ is approximately $61.07\ g$. Percent yield of the reaction is approximately $69.9\%$