The grams of aluminum that are required to produce 3.5 moles of AlO3 in presence of excess O2 is calculated as below
write the equation for reaction
4 Al + 3O2 =2 Al2O3
by use of mole ratio between Al to Al2O3 which is 4 :2 the moles of Al
=3.5 x4/2 = 7 moles
mass of Al = moles / x molar mass
= 7 moles x27 g/mol =189 grams
Answer:
phosphorous- 5
calcium- 2
nitrogen- 3 or 5
iron- 8 (transition metals use subshells as valence electrons)
argon- 8
potassium- 1
helium- 2
magnesium- 2
sulfur- 6
lithium- 1
iodine- 7
oxygen- 6
barium- 2
aluminum- 3
hydrogen- 1
xenon- 8
copper- 1
Source: my own chemistry notes
Answer:
A Li2O
Explanation:
Li has a 1+ charge and O has a 2- charge so to balance the charges there needs to be 2 Li for every 1 O
1.17×6×1023
Wich equals 7.02×1023
Your final answer is 7.02×1023
