11. Chromium is produced by reacting its oxide with aluminum. If 76g of Cr2O3 and 27g of Al completely react to form 51g of Al2O3, how many grams of Cr are formed?
This is a limiting reagent problem. Most are solved the same way. However, this one has a short cut you can use.
Write an equation and balance it.
Convert 51 g Al2O3 to moles. moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles Al2O3 to moles Cr.
Now convert moles Cr to grams. g = moles x molar mass.
response about what
To find the number of grams of Cr formed, we need to determine the stoichiometry of the reaction and use it to calculate the moles of Cr2O3 and Al.
First, let's write out the balanced equation for the reaction:
3 Cr2O3 + 2 Al -> 2 Al2O3 + 4 Cr
According to the equation, we need 3 moles of Cr2O3 to react with 2 moles of Al to produce 4 moles of Cr.
Next, let's calculate the number of moles of Cr2O3 and Al:
Molar mass of Cr2O3 = (2 * atomic mass of Cr) + (3 * atomic mass of O)
= (2 * 52 g/mol) + (3 * 16 g/mol)
= 152 g/mol
Molar mass of Al = atomic mass of Al
= 27 g/mol
Number of moles of Cr2O3 = Mass of Cr2O3 / Molar mass of Cr2O3
= 76 g / 152 g/mol
= 0.5 mol
Number of moles of Al = Mass of Al / Molar mass of Al
= 27 g / 27 g/mol
= 1 mol
Now, let's determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction, which results in the formation of the maximum possible amount of the product.
From the balanced equation, we can see that 3 moles of Cr2O3 react with 2 moles of Al to produce 4 moles of Cr. Therefore, the stoichiometry ratio of Cr2O3 to Al is 3:2, or 1.5:1.
However, in this case, we have 0.5 moles of Cr2O3 and 1 mole of Al. This means that 0.5 moles of Cr2O3 is equivalent to (1.5/1) * 0.5 = 0.75 moles of Al, which is less than the actual amount of Al available.
Therefore, Al is the limiting reactant.
Now, let's calculate the amount of Cr formed:
Number of moles of Cr = (Number of moles of Al) * (4 moles of Cr / 2 moles of Al)
= 1 mol * (4/2)
= 2 mol
Finally, let's calculate the mass of Cr:
Mass of Cr = Number of moles of Cr * Molar mass of Cr
= 2 mol * (52 g/mol)
= 104 g
Therefore, 104 grams of Cr are formed in the reaction.
To determine the amount of chromium formed, we need to solve the problem using stoichiometry. Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction.
Let's start by writing the balanced chemical equation:
2 Cr2O3 + 3 Al -> 4 Al2O3 + 4 Cr
According to the equation, 2 moles of Cr2O3 react with 3 moles of Al to produce 4 moles of Al2O3 and 4 moles of Cr.
First, we need to calculate the amount of moles for each substance involved:
Molar mass of Cr2O3 = (2 * atomic mass of Cr) + (3 * atomic mass of O)
= (2 * 52 g/mol) + (3 * 16 g/mol)
= 104 g/mol + 48 g/mol
= 152 g/mol
Molar mass of Al = 27 g/mol (given)
Molar mass of Al2O3 = (2 * atomic mass of Al) + (3 * atomic mass of O)
= (2 * 27 g/mol) + (3 * 16 g/mol)
= 54 g/mol + 48 g/mol
= 102 g/mol
Next, let's determine the moles of each substance:
Moles of Cr2O3 = Mass of Cr2O3 / Molar mass of Cr2O3
= 76 g / 152 g/mol
= 0.5 mol
Moles of Al = Mass of Al / Molar mass of Al
= 27 g / 27 g/mol
= 1 mol
Moles of Al2O3 = Mass of Al2O3 / Molar mass of Al2O3
= 51 g / 102 g/mol
= 0.5 mol
Since the molar ratio between Cr2O3 and Cr is 4:4, we can conclude that 0.5 moles of Cr2O3 will produce 0.5 moles of Cr.
Finally, we can calculate the mass of Cr formed:
Mass of Cr = Moles of Cr * Molar mass of Cr
= 0.5 mol * (atomic mass of Cr)
= 0.5 mol * 52 g/mol
= 26 g
Therefore, 26 grams of chromium (Cr) are formed in the reaction.