An element "E"E reacts with oxygen to form an oxide "E"_2"O"E2O. What is its relative atomic mass if 6.9 g of the element reacts with water to form "1800 cm"^31800 cm3 of hydrogen?

1 Answer
Ernest Z.
Apr 23, 2017

M_text(r) = 43Mr=43

Explanation:

"4E + 2O"_2 → "2E"_2"O"4E + 2O2→2E2O

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

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

Then, the reaction with water is

"2E + 2H"_2"O" → "2EOH" + "H"_22E + 2H2O→2EOH+H2

"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

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