# How much Chromium metal will produced from the above reaction?

1. Nov 2, 2008

### tiffgraf

1. The problem statement, all variables and given/known data
Chromium metal, Cr(s) can be prepared from reacting chromium oxide, Cr2O3(s), with aluminum, Al(s). Al2O3(s) is a by-product. Assume a reaction of 20,0 kg of chromium oxide with 5,00 kg of aluminum metal. 1)how much Chromium metal will produced from the above reaction? 2) who reagent is too much and how much is it?

2. Relevant equations

3. The attempt at a solution
one thought is that

Cr2O3 + 2 Al -> 2 Cr + Al2O3

M1= M(Cr2O3)= 152 g/mol
M2= M(Al)= 27 g/mol
-> n1 = 20E3/152 = 131.58 mol
-> n2= 5000/27 = 185.18 mol= number of mols Cr formed
mass of cr formed:m3=n2*M3 = 185.18 mol *52 g/mol= 9629 g

Now, according to stoichiometry n1/n2 = 1/2 ; n2=185.18 mol -> n1 = 92.59 mol (m1=n1*M1= 12222 g) so 7778 g Cr2O3 are in excess.

and one other is

Cr2O3(s) + 2 Al(s) = 2 Cr(s) + Al2O3(s)

2 Cr = 2 (52.00) = 104.00
3 O = 3 (16.00) = 48.00
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152.00

1000 g 20,0 kg Cr2O3(s) = 20,0 kg * (1000 kg / 1g) = 20.000g Cr2O3(s)
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1 kg
5,00 kg Al(s) = 5,00 kg * (1000 kg / 1g) = 5.000g Al

20.000 g Cr2O3(s) * 1 mol Cr2O3(s) / 152 g Cr2O3(s)* 2 mol Cr(s) / 1 mol Cr2O3(s) = 0.26314 mol Cr(s)

5.000 g Al(s) * 1 mol Al / 27 g Al(s) * 2 mol Cr / 2 mol Al(s) = 0.1851 mol Cr(s)

Cr2O3(s) is reactant excess

0.1851 mol Cr(s) * 1 mol Cr2O3(s) / 2 mol Cr(s) * 152g Cr2O3(s) / 1 mol Cr2O3(s) = 14.06 g

20.00 g - 14.06 g = 5.94 g Cr2O3(s) is how much excess left

0.1851 mol Cr * 52 g Cr(s) / 1 mol Cr = 9.6252 g Cr(s) was formed

2. Nov 2, 2008

### Staff: Mentor

Beware when using dot and comma in numbers, 20.000 is not the same as 20000.