Question

An element `"E"` reacts with oxygen to form an oxide `"E"_2"O"`. What is its relative atomic mass if 6.9 g of the element reacts with water to form `"1800 cm"^3` of hydrogen?

Answer 1

`M_text(r) = 43`

Explanation:

`"4E + 2O"_2 → "2E"_2"O"`

`"E"` is a Group 1 element because it takes two atoms of `"E"` to combine with one atom
of `"O"`.

The formula of its hydroxide must be `"EOH"`.

Then, the reaction with water is

`"2E + 2H"_2"O" → "2EOH" + "H"_2`

`"Moles of H"_2 = 1800 color(red)(cancel(color(black)("cm"^3 "H"_2))) × "1 mol H"_2/("22 400" color(red)(cancel(color(black)("cm"^3 "H"_2)))) = "0.0804 mol H"_2`

`"Moles of E" = 0.0804 color(red)(cancel(color(black)("mol H"_2))) × "2 mol E"/(1 color(red)(cancel(color(black)("mol H"_2)))) = "0.161 mol E"`

`"Molar mass of E" = "6.9 g"/"0.161 mol" = "43 g/mol"`

`A_text(r) = 43`