Question

`1 xx 10^21` molecules are removed from `280" mg"` of carbon monoxide. Calculate the number of moles of carbon monoxide left? [Answer:`8.4 xx 10^- 3" moles"`]

Answer 1

Approximately `8.4` millimoles

Explanation:

Total moles of Carbon monoxide (`"CO"`) is

`"n"_"Total" = (280 × 10^-3\ "g")/"28 g/mol" = 10^-2\ "mol"`

Moles of `"CO"` removed is

`"n"_"removed" = 1 × 10^21 cancel"molecules" × "1 mol"/(6.022 × 10^23 cancel"molecules")`

`color(white)("n"_"removed") = 1.66 × 10^-3\ "mol"`

Moles of `"CO"` left is

`"n"_"left" = "n"_"Total" - "n"_"removed"`

`color(white)("n"_"left") = 10^-2\ "mol" - (1.66 × 10^-3\ "mol")`

`color(white)("n"_"left") = (10 - 1.66) × 10^-3\ "mol"`

`color(white)("n"_"left") ≈ 8.4 × 10^-3\ "mol"`

Answer 2

`8,34*10^-3` moles

Explanation:

CO has a molecular mass of 28, so 280 mg are 10 millimoles.
`10^21` molecules are`10^21/N_A =1.66*10^-3` moles,
Subtracting `1.66*10^-3` moles from`10*10^-3` moles gives `8.34*10^-3` moles