Answer:
a) [tex] m_{Al} = 17.82 g [/tex]
b) [tex] m_{Cr} = 34.32 g [/tex]
c) [tex] m_{Cr} = 3.42 kg [/tex]
d) [tex] m_{Cr} = 3.42 tonnes [/tex]
Explanation:
The reaction is:
Cr₂O₃(s) + 2Al(s) → 2Cr(s) + Al₂O₃(s)
a) To find the Al mass needed to react with 50 g of Cr₂O₃, we need to calculate the number of moles of Cr₂O₃:
[tex] n_{Cr_{2}O_{3}} = \frac{m_{Cr_{2}O_{3}}}{M_{Cr_{2}O_{3}}} [/tex]
Where:
[tex]m_{Cr_{2}O_{3}}[/tex]: is the mass = 50 g
[tex]M_{Cr_{2}O_{3}}[/tex]: is the molar mass = 2*52+3*16 = 152 g/mol
[tex] n_{Cr_{2}O_{3}} = \frac{50 g}{152 g/mol} = 0.33 moles [/tex]
Now, the estoichiometric relation between Cr₂O₃ and Al is 1:2, so:
[tex] \eta_{Al} = \frac{2}{1}*\eta_{Cr_{2}O_{3}} = 2*0.33 moles = 0.66 moles [/tex]
Hence, the mass of Al is:
[tex] m_{Al} = 0.66 moles*27 g/mol = 17.82 g [/tex]
b) The stoichiometric relation from Cr₂O₃ and Cr is 1:2, hence:
[tex] \eta_{Cr} = \frac{2}{1}*0.33 moles = 0.66 moles [/tex]
Thus, the mass of Cr is:
[tex] m_{Cr} = 0.66 moles*52 g/mol = 34.32 g [/tex]
c) The number of moles of Cr₂O₃ with a mass of 5 kg is:
[tex] n_{Cr_{2}O_{3}} = \frac{5 \cdot 10^{3} g}{152 g/mol} = 32.89 moles [/tex]
So, the mass of Cr is:
[tex] m_{Cr} = 2*32.89 moles*52 g/mol = 3.42 kg [/tex]
d) The mass of Cr produced from 5 tonnes of Cr₂O₃ is:
[tex] m_{Cr} = 2*\frac{5 \cdot 10^{6} g}{152 g/mol}*52 g/mol = 3.42 tonnes [/tex]
I hope it helps you!
