How much Chromium metal will produced from the above reaction?

In summary, a reaction of 20.0 kg of chromium oxide and 5.00 kg of aluminum metal will produce 9.6252 g of chromium metal. The reagent in excess is chromium oxide, with 5.94 g remaining.
  • #1
tiffgraf
4
0

Homework Statement


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?


Homework Equations


please help me I am confused is the second solution right ?

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
------
152.00

1000 g 20,0 kg Cr2O3(s) = 20,0 kg * (1000 kg / 1g) = 20.000g Cr2O3(s)
-----------
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
 
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  • #2
tiffgraf said:
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)

Beware when using dot and comma in numbers, 20.000 is not the same as 20000.
 
  • #3


1) The above reaction will produce 9.6252 grams of chromium metal from the given amounts of chromium oxide and aluminum.

2) In this reaction, the reagent in excess is chromium oxide, with 5.94 grams of excess left over. This means that 5.94 grams of chromium oxide was not used in the reaction and therefore, 5.94 grams of aluminum was in excess.
 

1. What is the chemical equation for the production of Chromium metal?

The chemical equation for the production of Chromium metal is 2CrO3 + 3Al → 2Cr + 3Al2O3.

2. How much Chromium metal will be produced from this reaction?

The amount of Chromium metal produced from this reaction can be calculated using stoichiometry and the given amounts of CrO3 and Al.

3. What is the molar ratio of CrO3 to Cr in this reaction?

The molar ratio of CrO3 to Cr in this reaction is 2:2 or 1:1. This means that for every 2 moles of CrO3, 2 moles of Cr will be produced.

4. How does the amount of Al used in the reaction affect the production of Chromium metal?

The amount of Al used in the reaction directly affects the amount of Chromium metal produced. The more Al present, the more Cr will be produced, as Al is the reducing agent in this reaction.

5. Is this a balanced chemical equation?

Yes, this is a balanced chemical equation. All atoms present on the reactant side are also present on the product side, and the number of each type of atom is equal on both sides.

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