Question

04 question (2 points) open ebook section 3.5 chromium metal can be produced from the high temperature reaction of cr2o3 chromium(iii) oxide with silicon or aluminum by each of the following reactions: cr2o3 + 2al → 2cr + al2o3 2cr2o3 + 3si → 4cr + 3sio2 1st attempt part 1 (1 point) calculate the number of grams of aluminum required to prepare 331.0 g of chromium metal by the first reaction. g part 2 (1 point) calculate the number of grams of silicon required to prepare 331.0 g of chromium metal by the second reaction. g see hint

Explanation:

Step1: Determine molar masses

The molar mass of $Cr$ is $M_{Cr}=52.00\ g/mol$, the molar mass of $Al$ is $M_{Al} = 26.98\ g/mol$, and for $Si$, $M_{Si}=28.09\ g/mol$.

Step2: Calculate moles of chromium in Part 1

The number of moles of $Cr$ in $331.0\ g$ is $n_{Cr}=\frac{m_{Cr}}{M_{Cr}}=\frac{331.0\ g}{52.00\ g/mol}=6.365\ mol$.
From the reaction $Cr_{2}O_{3}+2Al
ightarrow2Cr + Al_{2}O_{3}$, the mole - ratio of $Al$ to $Cr$ is $1:1$. So the number of moles of $Al$ required, $n_{Al}=n_{Cr}=6.365\ mol$.
The mass of $Al$ is $m_{Al}=n_{Al}\times M_{Al}=6.365\ mol\times26.98\ g/mol = 171.7\ g$.

Step3: Calculate moles of chromium in Part 2

The number of moles of $Cr$ in $331.0\ g$ is still $n_{Cr}=\frac{331.0\ g}{52.00\ g/mol}=6.365\ mol$.
From the reaction $2Cr_{2}O_{3}+3Si
ightarrow4Cr + 3SiO_{2}$, the mole - ratio of $Si$ to $Cr$ is $\frac{3}{4}$.
So the number of moles of $Si$ required, $n_{Si}=\frac{3}{4}n_{Cr}=\frac{3}{4}\times6.365\ mol = 4.774\ mol$.
The mass of $Si$ is $m_{Si}=n_{Si}\times M_{Si}=4.774\ mol\times28.09\ g/mol=134.1\ g$.

Answer:

Part 1: 171.7
Part 2: 134.1