# 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?

Apr 23, 2017

${M}_{\textrm{r}} = 43$

#### Explanation:

$\text{4E + 2O"_2 → "2E"_2"O}$

$\text{E}$ is a Group 1 element because it takes two atoms of $\text{E}$ to combine with one atom
of $\text{O}$.

The formula of its hydroxide must be $\text{EOH}$.

Then, the reaction with water is

${\text{2E + 2H"_2"O" → "2EOH" + "H}}_{2}$

${\text{Moles of H"_2 = 1800 color(red)(cancel(color(black)("cm"^3 "H"_2))) × "1 mol H"_2/("22 400" color(red)(cancel(color(black)("cm"^3 "H"_2)))) = "0.0804 mol H}}_{2}$

$\text{Moles of E" = 0.0804 color(red)(cancel(color(black)("mol H"_2))) × "2 mol E"/(1 color(red)(cancel(color(black)("mol H"_2)))) = "0.161 mol E}$

$\text{Molar mass of E" = "6.9 g"/"0.161 mol" = "43 g/mol}$

${A}_{\textrm{r}} = 43$