if 33.0 grams of aluminium reacts with an excess of copper (II) chloride, what amount of pure copper will be p?
My periodic table lists Aluminium as 2.70 g / cm^3, which is 2700 kg / m^3, not 2800But, I digress, and do both for you: (8 kg) / (2800 kg / m^3) 0.002857 m^3 (8 kg) / (2700 kg / m^3) 0.002963 m^3 2857 cm^3 2963 cm^3 2857 cm^3 / 0.7 cm / pi / (2.5 cm)^2 207.8 holes 2963 cm^3 / 0.7 cm / pi / (2.5 cm)^2 215.6 holes If you start with 3200 kg per m^3, you'd get 182Sure it's not Barium or Scandium?
find moles, using molar mass 33.0 g Al 26.98 g/mol 1.223 moles of Al by the equation 2 Al 3 CuCl2 - 2 AlCl3 3 Cu 1.223 moles of Al produces 3/2 's as many moles of Cu 1.835 moles of Cu using molar mass 1.835 moles of Cu 63.55 g/mol 116.6 grams of Cu your answer rounded to 3 sig figs is 117 grams of Cu
I reviewed your work and it looks correct.
