Chromium is produced by reacting its oxide with aluminum. If 76 g of Cr2O3 and 27 g of Al completely react to form 51 g of Al2O3, how many grams of Cr are formed?
To find the amount of chromium (Cr) formed, we need to determine the amount of chromium oxide (Cr2O3) that reacted with aluminum (Al) to form aluminum oxide (Al2O3).
First, we calculate the molar mass of Cr2O3 and Al2O3:
- 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 Al2O3 = (2 * atomic mass of Al) + (3 * atomic mass of O)
= (2 * 27 g/mol) + (3 * 16 g/mol)
= 102 g/mol
Next, we need to calculate the number of moles of Al2O3 formed:
- Moles of Al2O3 = Mass of Al2O3 / Molar mass of Al2O3
= 51 g / 102 g/mol
= 0.5 mol
Since the reaction is balanced, we know that 1 mol of Cr2O3 reacts with 2 mol of Al to produce 1 mol of Al2O3 and 2 mol of Cr:
- Moles of Cr = 2 * Moles of Al2O3
= 2 * 0.5 mol
= 1 mol
Finally, we can calculate the mass of Cr formed using its molar mass:
- Mass of Cr = Moles of Cr * Molar mass of Cr
= 1 mol * (atomic mass of Cr)
= 1 * 52 g
= 52 g
Therefore, 52 grams of chromium (Cr) are formed in the reaction.