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?

1 Answer
Apr 23, 2017

Answer:

#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#