How many atoms of chlorine are there in 16.5 g of iron (iii) chloride?

Fecl3 has a RFM of 56 + (35.5 * 3) = 162.5

so find out its moles 16.5 ÷ 162 5 = aprox. 0.1M

then there is a 1 : 3 ratio so 0.1 * 3 = 0.3

then times that by the avogado constant which is 6.02×10^23

0.3 × 6.02×10^23 = ans

hope that helps