Question

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

Cr2O3(s) + 2Al(s) → 2Cr(s) + Al2O3(s)

a=Calculate the mass of aluminium needed to react with 50 g of Cr2O3.
b=Calculate the mass of chromium produced from 50 g of Cr2O3.
c=Calculate the mass of chromium produced from 5 kg of Cr2O3.
d=Calculate the mass of chromium produced from5 tonnes of Cr2O3.

Note: 1 tonne = 1,000,000 g
Ar: Cr = 52, O = 16, Al = 27

Answer 1

a)
`m_(Al) = 17.82 g`

b)
`m_(Cr) = 34.32 g`

c)
`m_(Cr) = 3.42 kg`

d)
`m_(Cr) = 3.42 tonnes`

Step-by-step 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)} = (50 g)/(152 g/mol) = 0.33 moles`

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

`\eta_(Al) = (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) = (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)} = (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*(5 \cdot 10^(6) g)/(152 g/mol)*52 g/mol = 3.42 tonnes`

I hope it helps you!