Answer :
The amount of Cr metal needed to produce 38.2 g of chromium(III) oxide is approximately 39.708 g, hence none of the options given are correct.
To solve this problem, we need to use the concept of stoichiometry and the balanced chemical equation for the formation of chromium(III) oxide ([tex]Cr_{2} O_{3}[/tex]).
The balanced chemical equation for the formation of chromium(III) oxide from chromium metal (Cr) is:
4 Cr + 3 [tex]O_{2}[/tex] → 2 [tex]Cr_{2} O_{3}[/tex]
From the equation, we can see that 4 moles of chromium (Cr) react with 3 moles of oxygen ([tex]O_{2}[/tex]) to produce 2 moles of chromium(III) oxide ([tex]Cr_{2} O_{3}[/tex]).
Now, let's calculate the molar mass of [tex]Cr_{2} O_{3}[/tex]:
Cr: 51.996 g/mol
O: 15.999 g/mol (x3) = 47.997 g/mol
Total molar mass of [tex]Cr_{2} O_{3}[/tex] = 51.996 + 47.997 = 99.993 g/mol
Next, let's calculate the number of moles of [tex]Cr_{2} O_{3}[/tex] produced:
38.2 g (mass of [tex]Cr_{2} O_{3}[/tex]) / 99.993 g/mol (molar mass of [tex]Cr_{2} O_{3}[/tex]) = 0.382 moles
According to the stoichiometry of the balanced equation, 4 moles of Cr produce 2 moles of [tex]Cr_{2} O_{3}[/tex].
So, to find the moles of Cr needed:
0.382 moles [tex]Cr_{2} O_{3}[/tex] × (4 moles Cr / 2 moles [tex]Cr_{2} O_{3}[/tex]) = 0.764 moles Cr
Finally, let's calculate the mass of Cr needed:
0.764 moles Cr × 51.996 g/mol (molar mass of Cr) = 39.708 g