4Al + 3O2 ---> 2Al2O3
When 1.75 moles of Al are reacted, how many grams of Al2O3 are also reacted?
To find the mass of Al2O3 reacted, you need to use the given number of moles of Al and the balanced chemical equation.
The balanced equation is 4Al + 3O2 -> 2Al2O3, which tells you that 4 moles of Al react to produce 2 moles of Al2O3.
Given that you have 1.75 moles of Al, you can set up a proportion using the mole ratio from the balanced equation:
4 moles of Al / 2 moles of Al2O3 = 1.75 moles of Al / x (grams of Al2O3)
Solving for x:
x = (1.75 moles of Al * 2 moles of Al2O3) / 4 moles of Al
x = 1.75 moles of Al2O3
Now, to convert moles of Al2O3 to grams, you need to know the molar mass. The molar mass of Al2O3 can be calculated by summing the atomic masses of aluminum (Al) and oxygen (O).
Al (atomic mass) = 26.98 g/mol
O (atomic mass) = 16.00 g/mol
2 Al atoms + 3 O atoms = (2 * 26.98 g/mol) + (3 * 16.00 g/mol) = 102.00 g/mol
So, to calculate the mass of Al2O3 in grams:
Mass = moles of Al2O3 * molar mass
Mass = 1.75 moles * 102.00 g/mol
Hence, when 1.75 moles of Al are reacted, the mass of Al2O3 reacted is 178.5 grams.