use the following balanced equation: 3ba + al2(so4)3 → 2al + 3baso4 if 12.0 g of ba react with 9.0 g of…

Answer:

B. $\ce{Al2(SO4)3}$

Explanation:

Step1: Calculate moles of $\ce{Ba}$

Molar - mass of $\ce{Ba}=137.33\ g/mol$. Moles of $\ce{Ba}=\frac{12.0\ g}{137.33\ g/mol}\approx0.0874\ mol$.

Step2: Calculate moles of $\ce{Al2(SO4)3}$

Molar - mass of $\ce{Al2(SO4)3}=2\times27 + 3\times(32 + 4\times16)=342.15\ g/mol$. Moles of $\ce{Al2(SO4)3}=\frac{9.0\ g}{342.15\ g/mol}\approx0.0263\ mol$.

Step3: Determine mole - ratio from the balanced equation

From $3\ce{Ba}+\ce{Al2(SO4)3}\rightarrow2\ce{Al}+3\ce{BaSO4}$, the mole - ratio of $\ce{Ba}$ to $\ce{Al2(SO4)3}$ is $3:1$.

Step4: Find the limiting and excess reactant

For $0.0263\ mol$ of $\ce{Al2(SO4)3}$, the moles of $\ce{Ba}$ required is $3\times0.0263\ mol = 0.0789\ mol$. Since we have $0.0874\ mol$ of $\ce{Ba}$ and only $0.0789\ mol$ is needed to react with $0.0263\ mol$ of $\ce{Al2(SO4)3}$, $\ce{Al2(SO4)3}$ is the limiting reactant and $\ce{Ba}$ is the excess reactant.