Question

given 4al + 3o₂ → 2al₂o₃. in this chemical reaction, how many grams of al₂o₃ will be produced if 375 grams of al react completely? express your answer to three significant figures. there will be grams of al₂o₃.

Explanation:

Step1: Determine molar masses

The molar mass of Al ($M_{Al}$) is approximately 26.98 g/mol, and the molar mass of $Al_2O_3$ ($M_{Al_2O_3}$) is $2\times26.98 + 3\times16.00= 101.96$ g/mol.

Step2: Calculate moles of Al

The number of moles of Al ($n_{Al}$) is calculated using the formula $n=\frac{m}{M}$, where $m = 375$ g and $M = M_{Al}=26.98$ g/mol. So, $n_{Al}=\frac{375}{26.98}\approx13.89$ mol.

Step3: Use stoichiometry

From the balanced chemical equation $4Al + 3O_2
ightarrow2Al_2O_3$, the mole - ratio of Al to $Al_2O_3$ is 4:2 or 2:1. So the number of moles of $Al_2O_3$ ($n_{Al_2O_3}$) produced is $n_{Al_2O_3}=\frac{1}{2}n_{Al}$. Substituting $n_{Al}\approx13.89$ mol, we get $n_{Al_2O_3}=\frac{13.89}{2}=6.945$ mol.

Step4: Calculate mass of $Al_2O_3$

The mass of $Al_2O_3$ ($m_{Al_2O_3}$) is calculated using the formula $m = n\times M$. Substituting $n = n_{Al_2O_3}=6.945$ mol and $M = M_{Al_2O_3}=101.96$ g/mol, we get $m_{Al_2O_3}=6.945\times101.96\approx708$ g.

Answer:

708 g