A pure titanium cube has an edge length of 2.63 in. How many titanium atoms does it contain? Titanium has a density of 4.50g/cm3. I messed something up along the line and would appreciate it if someone could tell me what I did wrong and how to solve it correctly. V = 2.63^3 = 18.19in^3
1in^3 = 16.387cm^3
18.19/16.387 = 1.1539
Density = 4.50g/cm^3
1.1539 * 4.50 = 5.1926
1mol Ti = g/47.867
5.1926/47.867 = 0.108479
1mol = 6.02 * 10^23atoms
0.108479/(6.02 * 10^23) = 6.53 * 10^22

This is wrong. Why!?

Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes