Question

Chromium is manufactured by heating a mixture of chromium(III) oxide with aluminium powder.

Answer 1

Answer:

a) $m_{Al} = 17.82 g$

b) $m_{Cr} = 34.32 g$

c) $m_{Cr} = 3.42 kg$

d) $m_{Cr} = 3.42 tonnes$            

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:

$n_{Cr_{2}O_{3}} = \frac{m_{Cr_{2}O_{3}}}{M_{Cr_{2}O_{3}}}$  

Where:    

$m_{Cr_{2}O_{3}}$: is the mass = 50 g

$M_{Cr_{2}O_{3}}$: is the molar mass = 2*52+3*16 = 152 g/mol

$n_{Cr_{2}O_{3}} = \frac{50 g}{152 g/mol} = 0.33 moles$

Now, the estoichiometric relation between Cr₂O₃ and Al is 1:2, so:

$\eta_{Al} = \frac{2}{1}*\eta_{Cr_{2}O_{3}} = 2*0.33 moles = 0.66 moles$  

Hence, the mass of  Al is:

$m_{Al} = 0.66 moles*27 g/mol = 17.82 g$

b) The stoichiometric relation from Cr₂O₃ and Cr is 1:2, hence:

$\eta_{Cr} = \frac{2}{1}*0.33 moles = 0.66 moles$

Thus, the mass of Cr is:

$m_{Cr} = 0.66 moles*52 g/mol = 34.32 g$

c) The number of moles of Cr₂O₃ with a mass of 5 kg is:

$n_{Cr_{2}O_{3}} = \frac{5 \cdot 10^{3} g}{152 g/mol} = 32.89 moles$        

So, the mass of Cr is:

$m_{Cr} = 2*32.89 moles*52 g/mol = 3.42 kg$

d) The mass of Cr produced from 5 tonnes of Cr₂O₃ is:

$m_{Cr} = 2*\frac{5 \cdot 10^{6} g}{152 g/mol}*52 g/mol = 3.42 tonnes$

I hope it helps you!