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 1×10²g 1.5×10²g a. what are the final masses of: i. the product ii. the reactants when the reaction has gone to completion? i) al = 1×10² / 26.98 ≈ 3.71 moles o₂ = 1.5×10² / 32 ≈ 4.69 moles

Explanation:

Step1: Calculate moles of reactants

Molar mass of Al is approximately $26.98\ g/mol$, so moles of Al, $n_{Al}=\frac{1.00\times 10^{2}\ g}{26.98\ g/mol}\approx3.71\ mol$. Molar mass of $O_{2}$ is $32\ g/mol$, so moles of $O_{2}$, $n_{O_{2}}=\frac{1.00\times 10^{2}\ g}{32\ g/mol}\approx 3.125\ mol$.

Step2: Determine limiting reactant

From the balanced equation $4Al + 3O_{2}
ightarrow2Al_{2}O_{3}$, the mole - ratio of Al to $O_{2}$ is $\frac{4}{3}$. For $3.71\ mol$ of Al, the moles of $O_{2}$ required is $n_{O_{2}\ required}=3.71\ mol\times\frac{3}{4}=2.7825\ mol$. Since we have $3.125\ mol$ of $O_{2}$, Al is the limiting reactant.

Step3: Calculate mass of product

From the balanced equation, 4 moles of Al produce 2 moles of $Al_{2}O_{3}$. Molar mass of $Al_{2}O_{3}$ is $2\times26.98 + 3\times16=101.96\ g/mol$. Since 4 moles of Al produce 2 moles of $Al_{2}O_{3}$, for $3.71\ mol$ of Al, moles of $Al_{2}O_{3}$ produced is $n_{Al_{2}O_{3}}=\frac{3.71\ mol\times2}{4}=1.855\ mol$. Mass of $Al_{2}O_{3}$, $m_{Al_{2}O_{3}}=1.855\ mol\times101.96\ g/mol\approx189\ g$.

Step4: Calculate remaining mass of non - limiting reactant

Moles of $O_{2}$ used is $2.7825\ mol$. Remaining moles of $O_{2}$ is $n_{O_{2}\ remaining}=3.125 - 2.7825 = 0.3425\ mol$. Mass of remaining $O_{2}$ is $m_{O_{2}\ remaining}=0.3425\ mol\times32\ g/mol = 10.96\ g$. Mass of Al remaining is 0 g as it is the limiting reactant and is completely consumed.

Answer:

i. Mass of product ($Al_{2}O_{3}$): 189 g
ii. Mass of Al: 0 g, Mass of $O_{2}$: 10.96 g