If 76g of Cr2O3 and 27g of Al completely ract to form 51g of Al2O3, how many grams of Cr are formed?
52 grams of 2Cr
To find out how many grams of Cr are formed, we need to determine the amount of Cr present in the reacted compounds.
The given information tells us that 76g of Cr2O3 and 27g of Al react to form 51g of Al2O3.
First, let's find the molar masses of the substances involved:
- Molar mass of Cr2O3 = (2 * atomic mass of Cr) + (3 * atomic mass of O)
= (2 * 52 g/mol) + (3 * 16 g/mol)
= 52 g/mol + 48 g/mol
= 100 g/mol
- 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, calculate the number of moles for each substance using their respective molar masses and given masses:
- Moles of Cr2O3 = 76g / (molar mass of Cr2O3)
= 76g / 100 g/mol
= 0.76 mol
- Moles of Al = 27g / (atomic mass of Al)
= 27g / 27 g/mol
= 1 mol
Now, let's compare the mole ratios of the reactants to determine which one is the limiting reactant. The balanced equation for the reaction is:
3 Cr2O3 + 4 Al -> 2 Al2O3 + 6 Cr
According to the balanced equation, the mole ratio of Cr2O3 to Cr is 6:3, which reduces to a 2:1 ratio.
Since we have 0.76 moles of Cr2O3, this means we could potentially produce twice as many moles of Cr. Therefore, we can expect a maximum of 0.76 * 2 = 1.52 mol of Cr to be formed.
Finally, let's find the mass of Cr using its molar mass:
Mass of Cr = (molar mass of Cr) * (mol of Cr)
= 52 g/mol * 1.52 mol
= 79.04 g
Therefore, approximately 79.04 grams of Cr are formed in the reaction.